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Maximum number of inputs per transaction

3

In theory, what is the maximum number of inputs per transaction? Is it defined primarily by the maximum number we can store on the 9 byte varint?

How big of a number can be stored on 9 bytes?

Would such a transaction even fit in a block if it had just 1 output?

transactions transaction-input input

share|improve this question

asked Apr 2 at 4:19

Pedro GonçalvesPedro Gonçalves

646

add a comment | 

3

In theory, what is the maximum number of inputs per transaction? Is it defined primarily by the maximum number we can store on the 9 byte varint?

How big of a number can be stored on 9 bytes?

Would such a transaction even fit in a block if it had just 1 output?

transactions transaction-input input

share|improve this question

asked Apr 2 at 4:19

Pedro GonçalvesPedro Gonçalves

646

add a comment | 

In theory, what is the maximum number of inputs per transaction? Is it defined primarily by the maximum number we can store on the 9 byte varint?

How big of a number can be stored on 9 bytes?

Would such a transaction even fit in a block if it had just 1 output?

transactions transaction-input input

share|improve this question

asked Apr 2 at 4:19

Pedro GonçalvesPedro Gonçalves

646

share|improve this question

asked Apr 2 at 4:19

Pedro GonçalvesPedro Gonçalves

646

share|improve this question

asked Apr 2 at 4:19

Pedro GonçalvesPedro Gonçalves

646

asked Apr 2 at 4:19

Pedro GonçalvesPedro Gonçalves

646

asked Apr 2 at 4:19

Pedro GonçalvesPedro Gonçalves

646

1 Answer
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3

No, the number that can be represented by the varint has no effect on the maximum number of inputs. That number is far too large. Rather the maximum number of inputs is constrained by the block size.

If it really matters to you what the maximum number that a varint can represent is, it's just the maximum value for a 64-bit integer. That's 0xffffffffffffffff. There's really no constraints on what a varint can represent. A transaction with such a number of inputs would not fit in a block, nor would it fit on any existing single storage medium as that transaction would be at least 664.1 Exabytes in size.

The maximum number of inputs that can fit in a valid transaction is 27022.

Such a transaction would not use segwit, so we use the maximum block size without segwit of 1000000 bytes. Subtract the 146 for the header and coinbase transaction to get 999854 bytes for the transaction. Subtract 4 bytes for the version, 4 bytes for the locktime, 1 byte for output count, 8 bytes for output value, 1 byte for output script, and 2 bytes for input count. This leaves us with 999833 bytes. With 37 bytes per input (32 previous txid, 4 output index, and 1 for script length), there can by 27022 inputs.

share|improve this answer

edited Apr 2 at 5:24

answered Apr 2 at 5:19

Andrew Chow♦Andrew Chow

33.7k42462

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  To note, that includes zero security or signatures.
  
  – Anonymous
  Apr 2 at 6:11
  

  
  
  
  

add a comment | 

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3

No, the number that can be represented by the varint has no effect on the maximum number of inputs. That number is far too large. Rather the maximum number of inputs is constrained by the block size.

If it really matters to you what the maximum number that a varint can represent is, it's just the maximum value for a 64-bit integer. That's 0xffffffffffffffff. There's really no constraints on what a varint can represent. A transaction with such a number of inputs would not fit in a block, nor would it fit on any existing single storage medium as that transaction would be at least 664.1 Exabytes in size.

The maximum number of inputs that can fit in a valid transaction is 27022.

Such a transaction would not use segwit, so we use the maximum block size without segwit of 1000000 bytes. Subtract the 146 for the header and coinbase transaction to get 999854 bytes for the transaction. Subtract 4 bytes for the version, 4 bytes for the locktime, 1 byte for output count, 8 bytes for output value, 1 byte for output script, and 2 bytes for input count. This leaves us with 999833 bytes. With 37 bytes per input (32 previous txid, 4 output index, and 1 for script length), there can by 27022 inputs.

share|improve this answer

edited Apr 2 at 5:24

answered Apr 2 at 5:19

Andrew Chow♦Andrew Chow

33.7k42462

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  To note, that includes zero security or signatures.
  
  – Anonymous
  Apr 2 at 6:11
  

  
  
  
  

add a comment | 

3

No, the number that can be represented by the varint has no effect on the maximum number of inputs. That number is far too large. Rather the maximum number of inputs is constrained by the block size.

If it really matters to you what the maximum number that a varint can represent is, it's just the maximum value for a 64-bit integer. That's 0xffffffffffffffff. There's really no constraints on what a varint can represent. A transaction with such a number of inputs would not fit in a block, nor would it fit on any existing single storage medium as that transaction would be at least 664.1 Exabytes in size.

The maximum number of inputs that can fit in a valid transaction is 27022.

Such a transaction would not use segwit, so we use the maximum block size without segwit of 1000000 bytes. Subtract the 146 for the header and coinbase transaction to get 999854 bytes for the transaction. Subtract 4 bytes for the version, 4 bytes for the locktime, 1 byte for output count, 8 bytes for output value, 1 byte for output script, and 2 bytes for input count. This leaves us with 999833 bytes. With 37 bytes per input (32 previous txid, 4 output index, and 1 for script length), there can by 27022 inputs.

share|improve this answer

edited Apr 2 at 5:24

answered Apr 2 at 5:19

Andrew Chow♦Andrew Chow

33.7k42462

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  To note, that includes zero security or signatures.
  
  – Anonymous
  Apr 2 at 6:11
  

  
  
  
  

add a comment | 

No, the number that can be represented by the varint has no effect on the maximum number of inputs. That number is far too large. Rather the maximum number of inputs is constrained by the block size.

If it really matters to you what the maximum number that a varint can represent is, it's just the maximum value for a 64-bit integer. That's 0xffffffffffffffff. There's really no constraints on what a varint can represent. A transaction with such a number of inputs would not fit in a block, nor would it fit on any existing single storage medium as that transaction would be at least 664.1 Exabytes in size.

The maximum number of inputs that can fit in a valid transaction is 27022.

Such a transaction would not use segwit, so we use the maximum block size without segwit of 1000000 bytes. Subtract the 146 for the header and coinbase transaction to get 999854 bytes for the transaction. Subtract 4 bytes for the version, 4 bytes for the locktime, 1 byte for output count, 8 bytes for output value, 1 byte for output script, and 2 bytes for input count. This leaves us with 999833 bytes. With 37 bytes per input (32 previous txid, 4 output index, and 1 for script length), there can by 27022 inputs.

share|improve this answer

edited Apr 2 at 5:24

answered Apr 2 at 5:19

Andrew Chow♦Andrew Chow

33.7k42462

share|improve this answer

edited Apr 2 at 5:24

answered Apr 2 at 5:19

Andrew Chow♦Andrew Chow

33.7k42462

answered Apr 2 at 5:19

Andrew Chow♦Andrew Chow

33.7k42462

answered Apr 2 at 5:19

Andrew Chow♦Andrew Chow

33.7k42462