Why does the compiler allow throws when the method will never throw the Exception The Next CEO of Stack OverflowHow slow are Java exceptions?The case against checked exceptionsWhy does Java have transient fields?Java: How to throw an Exception to the method caller inside a try catch body?Why does this code using random strings print “hello world”?Missing return statement in a non-void method compilesWhy does the Java compiler allow exceptions to be listed in the throws section that it is impossible for the method to throwWhy is “final” not allowed in Java 8 interface methods?Why is executing Java code in comments with certain Unicode characters allowed?Why is catching checked exceptions allowed for code that does not throw exceptions?
How exploitable/balanced is this homebrew spell: Spell Permanency?
Why did early computer designers eschew integers?
Could a dragon use its wings to swim?
Is it possible to create a QR code using text?
Why was Sir Cadogan fired?
Incomplete cube
Prodigo = pro + ago?
Does the Idaho Potato Commission associate potato skins with healthy eating?
Is it possible to make a 9x9 table fit within the default margins?
How to coordinate airplane tickets?
How to find if SQL server backup is encrypted with TDE without restoring the backup
Which acid/base does a strong base/acid react when added to a buffer solution?
Is it "common practice in Fourier transform spectroscopy to multiply the measured interferogram by an apodizing function"? If so, why?
Why does the freezing point matter when picking cooler ice packs?
Why did the Drakh emissary look so blurred in S04:E11 "Lines of Communication"?
How to compactly explain secondary and tertiary characters without resorting to stereotypes?
How can I separate the number from the unit in argument?
Is it a bad idea to plug the other end of ESD strap to wall ground?
Car headlights in a world without electricity
Why can't we say "I have been having a dog"?
Find a path from s to t using as few red nodes as possible
Are British MPs missing the point, with these 'Indicative Votes'?
How seriously should I take size and weight limits of hand luggage?
Gauss' Posthumous Publications?
Why does the compiler allow throws when the method will never throw the Exception
The Next CEO of Stack OverflowHow slow are Java exceptions?The case against checked exceptionsWhy does Java have transient fields?Java: How to throw an Exception to the method caller inside a try catch body?Why does this code using random strings print “hello world”?Missing return statement in a non-void method compilesWhy does the Java compiler allow exceptions to be listed in the throws section that it is impossible for the method to throwWhy is “final” not allowed in Java 8 interface methods?Why is executing Java code in comments with certain Unicode characters allowed?Why is catching checked exceptions allowed for code that does not throw exceptions?
I am wondering why the java compiler allows throws in the method declaration when the method will never throw the Exception. Because "throws" is a way of handling the exception (telling the caller to handle it).
Since there are two ways of handling exception (throws & try/catch). In a try/catch, it doesn't allow the catch of an exception not thrown in the try block but it allows a throws in a method that does may not throw the exception.
private static void methodA()
try
// Do something
// No IO operation here
catch (IOException ex) //This line does not compile because
//exception is never thrown from try
// Handle
private static void methodB() throws IOException //Why does this //compile when excetion is never thrown in function body
//Do Something
//No IO operation
java checked-exceptions
add a comment |
I am wondering why the java compiler allows throws in the method declaration when the method will never throw the Exception. Because "throws" is a way of handling the exception (telling the caller to handle it).
Since there are two ways of handling exception (throws & try/catch). In a try/catch, it doesn't allow the catch of an exception not thrown in the try block but it allows a throws in a method that does may not throw the exception.
private static void methodA()
try
// Do something
// No IO operation here
catch (IOException ex) //This line does not compile because
//exception is never thrown from try
// Handle
private static void methodB() throws IOException //Why does this //compile when excetion is never thrown in function body
//Do Something
//No IO operation
java checked-exceptions
2
TLDR: it's a design choice, and there isn't right or wrong here.
– yaseco
Mar 25 at 9:44
1
As we are not the designers of the language, "why" questions would be hard to answer with anything but speculative, opinion-based answers.
– RealSkeptic
Mar 25 at 9:45
I also find the termthrows
to be misleading. I parse it ascould_throw
in my mind now and it seems to fit better with its behaviour.
– Eric Duminil
Mar 25 at 15:54
add a comment |
I am wondering why the java compiler allows throws in the method declaration when the method will never throw the Exception. Because "throws" is a way of handling the exception (telling the caller to handle it).
Since there are two ways of handling exception (throws & try/catch). In a try/catch, it doesn't allow the catch of an exception not thrown in the try block but it allows a throws in a method that does may not throw the exception.
private static void methodA()
try
// Do something
// No IO operation here
catch (IOException ex) //This line does not compile because
//exception is never thrown from try
// Handle
private static void methodB() throws IOException //Why does this //compile when excetion is never thrown in function body
//Do Something
//No IO operation
java checked-exceptions
I am wondering why the java compiler allows throws in the method declaration when the method will never throw the Exception. Because "throws" is a way of handling the exception (telling the caller to handle it).
Since there are two ways of handling exception (throws & try/catch). In a try/catch, it doesn't allow the catch of an exception not thrown in the try block but it allows a throws in a method that does may not throw the exception.
private static void methodA()
try
// Do something
// No IO operation here
catch (IOException ex) //This line does not compile because
//exception is never thrown from try
// Handle
private static void methodB() throws IOException //Why does this //compile when excetion is never thrown in function body
//Do Something
//No IO operation
java checked-exceptions
java checked-exceptions
asked Mar 25 at 9:38
PsyApPsyAp
785
785
2
TLDR: it's a design choice, and there isn't right or wrong here.
– yaseco
Mar 25 at 9:44
1
As we are not the designers of the language, "why" questions would be hard to answer with anything but speculative, opinion-based answers.
– RealSkeptic
Mar 25 at 9:45
I also find the termthrows
to be misleading. I parse it ascould_throw
in my mind now and it seems to fit better with its behaviour.
– Eric Duminil
Mar 25 at 15:54
add a comment |
2
TLDR: it's a design choice, and there isn't right or wrong here.
– yaseco
Mar 25 at 9:44
1
As we are not the designers of the language, "why" questions would be hard to answer with anything but speculative, opinion-based answers.
– RealSkeptic
Mar 25 at 9:45
I also find the termthrows
to be misleading. I parse it ascould_throw
in my mind now and it seems to fit better with its behaviour.
– Eric Duminil
Mar 25 at 15:54
2
2
TLDR: it's a design choice, and there isn't right or wrong here.
– yaseco
Mar 25 at 9:44
TLDR: it's a design choice, and there isn't right or wrong here.
– yaseco
Mar 25 at 9:44
1
1
As we are not the designers of the language, "why" questions would be hard to answer with anything but speculative, opinion-based answers.
– RealSkeptic
Mar 25 at 9:45
As we are not the designers of the language, "why" questions would be hard to answer with anything but speculative, opinion-based answers.
– RealSkeptic
Mar 25 at 9:45
I also find the term
throws
to be misleading. I parse it as could_throw
in my mind now and it seems to fit better with its behaviour.– Eric Duminil
Mar 25 at 15:54
I also find the term
throws
to be misleading. I parse it as could_throw
in my mind now and it seems to fit better with its behaviour.– Eric Duminil
Mar 25 at 15:54
add a comment |
4 Answers
4
active
oldest
votes
The throws
clause is part of the method's contract. It requires the caller of the method to behave as if the specified exception may be thrown by the method (i.e. either catch the exception or declare their own throws
clause).
It's possible that the initial version of a method does not throw the exception specified in the throws
clause, but a future version can throw it without breaking the API (i.e. any existing code that calls the method will still pass compilation).
The opposite it also possible. If the method used to throw the exception specified in the throws
clause, but a future version of it doesn't throw it anymore, you should keep the throws
clause in order not to break existing code that uses your method.
First example:
Suppose you have this code which uses methodB
:
private static void methodA()
methodB(); // doesn't have throws IOException clause yet
If later you want to change methodB
to throw IOException
, methodA
will stop passing compilation.
Second example:
Suppose you have this code which uses methodB
:
private static void methodA()
try
methodB(); // throws IOException
catch (IOException ex)
If you remove the throws
clause from a future version of methodB
, methodA
won't pass compilation anymore.
This example is not very interesting when methodA
is private
, because it can only be used locally (within the same class, where it's easy to modify all the methods that call it).
However, if it becomes public
, you don't know who uses (or will use) your method, so you have no control of all the code that may break as a result of adding or removing the throws
clause.
And if it's an instance method, there's another reason for allowing the throws
clause even if you don't throw the exception - the method can be overridden, and the overriding method may throw the exception even if the base class implementation does not.
2
The last point about the override should be given more emphasis, IMHO. The notion of "what if functions change" is less concrete than the notion of "even though the parent function would have no reason to throw a particular exceptions, even today's versions of child functions might need to do so.
– supercat
Mar 25 at 17:47
@supercat agreed, but the code in the question has static methods, which cannot be overridden, which is why most my answer gives reasons that apply to static methods.
– Eran
Mar 25 at 19:42
add a comment |
Because the signature defines the contract of the method. Even if the method doesn't throw an IOException now, maybe it will in the future, and you want to prepare for that possibility.
Suppose you just provide a dummy implementation for the method for now, but you know that, later, the actual implementation will potentially throw an IOException.
If the compiler prevented you from adding this throws clause, you would be forced to rework all the calls (recursively) to that method once you provide the actual implementation of the method.
1
It's debatable whether a private method can be said to have a contract.
– RealSkeptic
Mar 25 at 9:44
1
It defines a contract for all the other methods in the class. And these methods might, in turn,want to declare an IOException because they rely on the contract of the private method they call.
– JB Nizet
Mar 25 at 9:47
You can also decide to change the return value of a private method. That's also part of the "contract", isn't it? And yet, what you do when that happens is fix the code that calls it. The same could have been done with "throws".
– RealSkeptic
Mar 25 at 9:49
I don't see how that relates to the question, which is: why does the compiler allow to have a throws clause for an exception that is not thrown by the method. What's your point? The compiler also allows you to uselong
as the return type even if you only ever return constant values that fit into an int. Why? For the same reason: you define the contract, which is that the method may, now or later, return a long value.
– JB Nizet
Mar 25 at 9:54
add a comment |
Keyword throws tells programmer that there could be an IOException happening in the method. Now if you didn't specify try/catch it means that when exception is thrown program will stop working, while in try/catch you handle it by doing something else if thrown exception.
Use throws for readability and specifying the possibility of exception and use try/catch to tell program what to do in case of exception.
add a comment |
The methodB throws IOException, so the method calling methodB is responsible for catching the exception which will be thrown by methodB. Try calling methodB from other methods, it will ask you to catch it or re-throw the IOException. Somewhere you will have to catch the IOException in the chain (in try/catch block). So you wont get compile time error.
private void sampleMethod()
try
methodB();
catch (IOException e)
// TODO Auto-generated catch block
e.printStackTrace();
try/catch in methodA admittedly swallows the exception, that means methodA is responsible for catching the exception in try/catch block.
"Every statement in any java program must be reachable i.e every statement must be executable at least once"
So, you will get compiler error as your try block doesn't have any code to cause IOException.
add a comment |
StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55334879%2fwhy-does-the-compiler-allow-throws-when-the-method-will-never-throw-the-exceptio%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
The throws
clause is part of the method's contract. It requires the caller of the method to behave as if the specified exception may be thrown by the method (i.e. either catch the exception or declare their own throws
clause).
It's possible that the initial version of a method does not throw the exception specified in the throws
clause, but a future version can throw it without breaking the API (i.e. any existing code that calls the method will still pass compilation).
The opposite it also possible. If the method used to throw the exception specified in the throws
clause, but a future version of it doesn't throw it anymore, you should keep the throws
clause in order not to break existing code that uses your method.
First example:
Suppose you have this code which uses methodB
:
private static void methodA()
methodB(); // doesn't have throws IOException clause yet
If later you want to change methodB
to throw IOException
, methodA
will stop passing compilation.
Second example:
Suppose you have this code which uses methodB
:
private static void methodA()
try
methodB(); // throws IOException
catch (IOException ex)
If you remove the throws
clause from a future version of methodB
, methodA
won't pass compilation anymore.
This example is not very interesting when methodA
is private
, because it can only be used locally (within the same class, where it's easy to modify all the methods that call it).
However, if it becomes public
, you don't know who uses (or will use) your method, so you have no control of all the code that may break as a result of adding or removing the throws
clause.
And if it's an instance method, there's another reason for allowing the throws
clause even if you don't throw the exception - the method can be overridden, and the overriding method may throw the exception even if the base class implementation does not.
2
The last point about the override should be given more emphasis, IMHO. The notion of "what if functions change" is less concrete than the notion of "even though the parent function would have no reason to throw a particular exceptions, even today's versions of child functions might need to do so.
– supercat
Mar 25 at 17:47
@supercat agreed, but the code in the question has static methods, which cannot be overridden, which is why most my answer gives reasons that apply to static methods.
– Eran
Mar 25 at 19:42
add a comment |
The throws
clause is part of the method's contract. It requires the caller of the method to behave as if the specified exception may be thrown by the method (i.e. either catch the exception or declare their own throws
clause).
It's possible that the initial version of a method does not throw the exception specified in the throws
clause, but a future version can throw it without breaking the API (i.e. any existing code that calls the method will still pass compilation).
The opposite it also possible. If the method used to throw the exception specified in the throws
clause, but a future version of it doesn't throw it anymore, you should keep the throws
clause in order not to break existing code that uses your method.
First example:
Suppose you have this code which uses methodB
:
private static void methodA()
methodB(); // doesn't have throws IOException clause yet
If later you want to change methodB
to throw IOException
, methodA
will stop passing compilation.
Second example:
Suppose you have this code which uses methodB
:
private static void methodA()
try
methodB(); // throws IOException
catch (IOException ex)
If you remove the throws
clause from a future version of methodB
, methodA
won't pass compilation anymore.
This example is not very interesting when methodA
is private
, because it can only be used locally (within the same class, where it's easy to modify all the methods that call it).
However, if it becomes public
, you don't know who uses (or will use) your method, so you have no control of all the code that may break as a result of adding or removing the throws
clause.
And if it's an instance method, there's another reason for allowing the throws
clause even if you don't throw the exception - the method can be overridden, and the overriding method may throw the exception even if the base class implementation does not.
2
The last point about the override should be given more emphasis, IMHO. The notion of "what if functions change" is less concrete than the notion of "even though the parent function would have no reason to throw a particular exceptions, even today's versions of child functions might need to do so.
– supercat
Mar 25 at 17:47
@supercat agreed, but the code in the question has static methods, which cannot be overridden, which is why most my answer gives reasons that apply to static methods.
– Eran
Mar 25 at 19:42
add a comment |
The throws
clause is part of the method's contract. It requires the caller of the method to behave as if the specified exception may be thrown by the method (i.e. either catch the exception or declare their own throws
clause).
It's possible that the initial version of a method does not throw the exception specified in the throws
clause, but a future version can throw it without breaking the API (i.e. any existing code that calls the method will still pass compilation).
The opposite it also possible. If the method used to throw the exception specified in the throws
clause, but a future version of it doesn't throw it anymore, you should keep the throws
clause in order not to break existing code that uses your method.
First example:
Suppose you have this code which uses methodB
:
private static void methodA()
methodB(); // doesn't have throws IOException clause yet
If later you want to change methodB
to throw IOException
, methodA
will stop passing compilation.
Second example:
Suppose you have this code which uses methodB
:
private static void methodA()
try
methodB(); // throws IOException
catch (IOException ex)
If you remove the throws
clause from a future version of methodB
, methodA
won't pass compilation anymore.
This example is not very interesting when methodA
is private
, because it can only be used locally (within the same class, where it's easy to modify all the methods that call it).
However, if it becomes public
, you don't know who uses (or will use) your method, so you have no control of all the code that may break as a result of adding or removing the throws
clause.
And if it's an instance method, there's another reason for allowing the throws
clause even if you don't throw the exception - the method can be overridden, and the overriding method may throw the exception even if the base class implementation does not.
The throws
clause is part of the method's contract. It requires the caller of the method to behave as if the specified exception may be thrown by the method (i.e. either catch the exception or declare their own throws
clause).
It's possible that the initial version of a method does not throw the exception specified in the throws
clause, but a future version can throw it without breaking the API (i.e. any existing code that calls the method will still pass compilation).
The opposite it also possible. If the method used to throw the exception specified in the throws
clause, but a future version of it doesn't throw it anymore, you should keep the throws
clause in order not to break existing code that uses your method.
First example:
Suppose you have this code which uses methodB
:
private static void methodA()
methodB(); // doesn't have throws IOException clause yet
If later you want to change methodB
to throw IOException
, methodA
will stop passing compilation.
Second example:
Suppose you have this code which uses methodB
:
private static void methodA()
try
methodB(); // throws IOException
catch (IOException ex)
If you remove the throws
clause from a future version of methodB
, methodA
won't pass compilation anymore.
This example is not very interesting when methodA
is private
, because it can only be used locally (within the same class, where it's easy to modify all the methods that call it).
However, if it becomes public
, you don't know who uses (or will use) your method, so you have no control of all the code that may break as a result of adding or removing the throws
clause.
And if it's an instance method, there's another reason for allowing the throws
clause even if you don't throw the exception - the method can be overridden, and the overriding method may throw the exception even if the base class implementation does not.
edited Mar 25 at 9:53
answered Mar 25 at 9:43
EranEran
291k37479563
291k37479563
2
The last point about the override should be given more emphasis, IMHO. The notion of "what if functions change" is less concrete than the notion of "even though the parent function would have no reason to throw a particular exceptions, even today's versions of child functions might need to do so.
– supercat
Mar 25 at 17:47
@supercat agreed, but the code in the question has static methods, which cannot be overridden, which is why most my answer gives reasons that apply to static methods.
– Eran
Mar 25 at 19:42
add a comment |
2
The last point about the override should be given more emphasis, IMHO. The notion of "what if functions change" is less concrete than the notion of "even though the parent function would have no reason to throw a particular exceptions, even today's versions of child functions might need to do so.
– supercat
Mar 25 at 17:47
@supercat agreed, but the code in the question has static methods, which cannot be overridden, which is why most my answer gives reasons that apply to static methods.
– Eran
Mar 25 at 19:42
2
2
The last point about the override should be given more emphasis, IMHO. The notion of "what if functions change" is less concrete than the notion of "even though the parent function would have no reason to throw a particular exceptions, even today's versions of child functions might need to do so.
– supercat
Mar 25 at 17:47
The last point about the override should be given more emphasis, IMHO. The notion of "what if functions change" is less concrete than the notion of "even though the parent function would have no reason to throw a particular exceptions, even today's versions of child functions might need to do so.
– supercat
Mar 25 at 17:47
@supercat agreed, but the code in the question has static methods, which cannot be overridden, which is why most my answer gives reasons that apply to static methods.
– Eran
Mar 25 at 19:42
@supercat agreed, but the code in the question has static methods, which cannot be overridden, which is why most my answer gives reasons that apply to static methods.
– Eran
Mar 25 at 19:42
add a comment |
Because the signature defines the contract of the method. Even if the method doesn't throw an IOException now, maybe it will in the future, and you want to prepare for that possibility.
Suppose you just provide a dummy implementation for the method for now, but you know that, later, the actual implementation will potentially throw an IOException.
If the compiler prevented you from adding this throws clause, you would be forced to rework all the calls (recursively) to that method once you provide the actual implementation of the method.
1
It's debatable whether a private method can be said to have a contract.
– RealSkeptic
Mar 25 at 9:44
1
It defines a contract for all the other methods in the class. And these methods might, in turn,want to declare an IOException because they rely on the contract of the private method they call.
– JB Nizet
Mar 25 at 9:47
You can also decide to change the return value of a private method. That's also part of the "contract", isn't it? And yet, what you do when that happens is fix the code that calls it. The same could have been done with "throws".
– RealSkeptic
Mar 25 at 9:49
I don't see how that relates to the question, which is: why does the compiler allow to have a throws clause for an exception that is not thrown by the method. What's your point? The compiler also allows you to uselong
as the return type even if you only ever return constant values that fit into an int. Why? For the same reason: you define the contract, which is that the method may, now or later, return a long value.
– JB Nizet
Mar 25 at 9:54
add a comment |
Because the signature defines the contract of the method. Even if the method doesn't throw an IOException now, maybe it will in the future, and you want to prepare for that possibility.
Suppose you just provide a dummy implementation for the method for now, but you know that, later, the actual implementation will potentially throw an IOException.
If the compiler prevented you from adding this throws clause, you would be forced to rework all the calls (recursively) to that method once you provide the actual implementation of the method.
1
It's debatable whether a private method can be said to have a contract.
– RealSkeptic
Mar 25 at 9:44
1
It defines a contract for all the other methods in the class. And these methods might, in turn,want to declare an IOException because they rely on the contract of the private method they call.
– JB Nizet
Mar 25 at 9:47
You can also decide to change the return value of a private method. That's also part of the "contract", isn't it? And yet, what you do when that happens is fix the code that calls it. The same could have been done with "throws".
– RealSkeptic
Mar 25 at 9:49
I don't see how that relates to the question, which is: why does the compiler allow to have a throws clause for an exception that is not thrown by the method. What's your point? The compiler also allows you to uselong
as the return type even if you only ever return constant values that fit into an int. Why? For the same reason: you define the contract, which is that the method may, now or later, return a long value.
– JB Nizet
Mar 25 at 9:54
add a comment |
Because the signature defines the contract of the method. Even if the method doesn't throw an IOException now, maybe it will in the future, and you want to prepare for that possibility.
Suppose you just provide a dummy implementation for the method for now, but you know that, later, the actual implementation will potentially throw an IOException.
If the compiler prevented you from adding this throws clause, you would be forced to rework all the calls (recursively) to that method once you provide the actual implementation of the method.
Because the signature defines the contract of the method. Even if the method doesn't throw an IOException now, maybe it will in the future, and you want to prepare for that possibility.
Suppose you just provide a dummy implementation for the method for now, but you know that, later, the actual implementation will potentially throw an IOException.
If the compiler prevented you from adding this throws clause, you would be forced to rework all the calls (recursively) to that method once you provide the actual implementation of the method.
answered Mar 25 at 9:43
JB NizetJB Nizet
547k588941021
547k588941021
1
It's debatable whether a private method can be said to have a contract.
– RealSkeptic
Mar 25 at 9:44
1
It defines a contract for all the other methods in the class. And these methods might, in turn,want to declare an IOException because they rely on the contract of the private method they call.
– JB Nizet
Mar 25 at 9:47
You can also decide to change the return value of a private method. That's also part of the "contract", isn't it? And yet, what you do when that happens is fix the code that calls it. The same could have been done with "throws".
– RealSkeptic
Mar 25 at 9:49
I don't see how that relates to the question, which is: why does the compiler allow to have a throws clause for an exception that is not thrown by the method. What's your point? The compiler also allows you to uselong
as the return type even if you only ever return constant values that fit into an int. Why? For the same reason: you define the contract, which is that the method may, now or later, return a long value.
– JB Nizet
Mar 25 at 9:54
add a comment |
1
It's debatable whether a private method can be said to have a contract.
– RealSkeptic
Mar 25 at 9:44
1
It defines a contract for all the other methods in the class. And these methods might, in turn,want to declare an IOException because they rely on the contract of the private method they call.
– JB Nizet
Mar 25 at 9:47
You can also decide to change the return value of a private method. That's also part of the "contract", isn't it? And yet, what you do when that happens is fix the code that calls it. The same could have been done with "throws".
– RealSkeptic
Mar 25 at 9:49
I don't see how that relates to the question, which is: why does the compiler allow to have a throws clause for an exception that is not thrown by the method. What's your point? The compiler also allows you to uselong
as the return type even if you only ever return constant values that fit into an int. Why? For the same reason: you define the contract, which is that the method may, now or later, return a long value.
– JB Nizet
Mar 25 at 9:54
1
1
It's debatable whether a private method can be said to have a contract.
– RealSkeptic
Mar 25 at 9:44
It's debatable whether a private method can be said to have a contract.
– RealSkeptic
Mar 25 at 9:44
1
1
It defines a contract for all the other methods in the class. And these methods might, in turn,want to declare an IOException because they rely on the contract of the private method they call.
– JB Nizet
Mar 25 at 9:47
It defines a contract for all the other methods in the class. And these methods might, in turn,want to declare an IOException because they rely on the contract of the private method they call.
– JB Nizet
Mar 25 at 9:47
You can also decide to change the return value of a private method. That's also part of the "contract", isn't it? And yet, what you do when that happens is fix the code that calls it. The same could have been done with "throws".
– RealSkeptic
Mar 25 at 9:49
You can also decide to change the return value of a private method. That's also part of the "contract", isn't it? And yet, what you do when that happens is fix the code that calls it. The same could have been done with "throws".
– RealSkeptic
Mar 25 at 9:49
I don't see how that relates to the question, which is: why does the compiler allow to have a throws clause for an exception that is not thrown by the method. What's your point? The compiler also allows you to use
long
as the return type even if you only ever return constant values that fit into an int. Why? For the same reason: you define the contract, which is that the method may, now or later, return a long value.– JB Nizet
Mar 25 at 9:54
I don't see how that relates to the question, which is: why does the compiler allow to have a throws clause for an exception that is not thrown by the method. What's your point? The compiler also allows you to use
long
as the return type even if you only ever return constant values that fit into an int. Why? For the same reason: you define the contract, which is that the method may, now or later, return a long value.– JB Nizet
Mar 25 at 9:54
add a comment |
Keyword throws tells programmer that there could be an IOException happening in the method. Now if you didn't specify try/catch it means that when exception is thrown program will stop working, while in try/catch you handle it by doing something else if thrown exception.
Use throws for readability and specifying the possibility of exception and use try/catch to tell program what to do in case of exception.
add a comment |
Keyword throws tells programmer that there could be an IOException happening in the method. Now if you didn't specify try/catch it means that when exception is thrown program will stop working, while in try/catch you handle it by doing something else if thrown exception.
Use throws for readability and specifying the possibility of exception and use try/catch to tell program what to do in case of exception.
add a comment |
Keyword throws tells programmer that there could be an IOException happening in the method. Now if you didn't specify try/catch it means that when exception is thrown program will stop working, while in try/catch you handle it by doing something else if thrown exception.
Use throws for readability and specifying the possibility of exception and use try/catch to tell program what to do in case of exception.
Keyword throws tells programmer that there could be an IOException happening in the method. Now if you didn't specify try/catch it means that when exception is thrown program will stop working, while in try/catch you handle it by doing something else if thrown exception.
Use throws for readability and specifying the possibility of exception and use try/catch to tell program what to do in case of exception.
answered Mar 25 at 9:50
IvanMIvanM
423
423
add a comment |
add a comment |
The methodB throws IOException, so the method calling methodB is responsible for catching the exception which will be thrown by methodB. Try calling methodB from other methods, it will ask you to catch it or re-throw the IOException. Somewhere you will have to catch the IOException in the chain (in try/catch block). So you wont get compile time error.
private void sampleMethod()
try
methodB();
catch (IOException e)
// TODO Auto-generated catch block
e.printStackTrace();
try/catch in methodA admittedly swallows the exception, that means methodA is responsible for catching the exception in try/catch block.
"Every statement in any java program must be reachable i.e every statement must be executable at least once"
So, you will get compiler error as your try block doesn't have any code to cause IOException.
add a comment |
The methodB throws IOException, so the method calling methodB is responsible for catching the exception which will be thrown by methodB. Try calling methodB from other methods, it will ask you to catch it or re-throw the IOException. Somewhere you will have to catch the IOException in the chain (in try/catch block). So you wont get compile time error.
private void sampleMethod()
try
methodB();
catch (IOException e)
// TODO Auto-generated catch block
e.printStackTrace();
try/catch in methodA admittedly swallows the exception, that means methodA is responsible for catching the exception in try/catch block.
"Every statement in any java program must be reachable i.e every statement must be executable at least once"
So, you will get compiler error as your try block doesn't have any code to cause IOException.
add a comment |
The methodB throws IOException, so the method calling methodB is responsible for catching the exception which will be thrown by methodB. Try calling methodB from other methods, it will ask you to catch it or re-throw the IOException. Somewhere you will have to catch the IOException in the chain (in try/catch block). So you wont get compile time error.
private void sampleMethod()
try
methodB();
catch (IOException e)
// TODO Auto-generated catch block
e.printStackTrace();
try/catch in methodA admittedly swallows the exception, that means methodA is responsible for catching the exception in try/catch block.
"Every statement in any java program must be reachable i.e every statement must be executable at least once"
So, you will get compiler error as your try block doesn't have any code to cause IOException.
The methodB throws IOException, so the method calling methodB is responsible for catching the exception which will be thrown by methodB. Try calling methodB from other methods, it will ask you to catch it or re-throw the IOException. Somewhere you will have to catch the IOException in the chain (in try/catch block). So you wont get compile time error.
private void sampleMethod()
try
methodB();
catch (IOException e)
// TODO Auto-generated catch block
e.printStackTrace();
try/catch in methodA admittedly swallows the exception, that means methodA is responsible for catching the exception in try/catch block.
"Every statement in any java program must be reachable i.e every statement must be executable at least once"
So, you will get compiler error as your try block doesn't have any code to cause IOException.
answered Mar 25 at 10:49
Harsimran17Harsimran17
92
92
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55334879%2fwhy-does-the-compiler-allow-throws-when-the-method-will-never-throw-the-exceptio%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
2
TLDR: it's a design choice, and there isn't right or wrong here.
– yaseco
Mar 25 at 9:44
1
As we are not the designers of the language, "why" questions would be hard to answer with anything but speculative, opinion-based answers.
– RealSkeptic
Mar 25 at 9:45
I also find the term
throws
to be misleading. I parse it ascould_throw
in my mind now and it seems to fit better with its behaviour.– Eric Duminil
Mar 25 at 15:54