Maximum number of inputs per transactionCan we interpret a rolling average transaction count per block as the 'adoption curve' for Bitcoin?Recommended Transaction SizeOptimising the inputs for a transactionWhat is the maximum number of output addresses I can send to with one bitcoin transaction?Get transaction fees per transaction via gettransactionWhat would be the implications of limiting Bitcoin transactions to fifty inputs?Determine inputs/outputs of a transaction before sendingIs there a maximum fee per byte in satoshis?Why does Bitcoin need transaction fees prior to hitting its maximum coin limit?What happens if “transaction size” is larger than “maximum block size”?
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Maximum number of inputs per transaction
Can we interpret a rolling average transaction count per block as the 'adoption curve' for Bitcoin?Recommended Transaction SizeOptimising the inputs for a transactionWhat is the maximum number of output addresses I can send to with one bitcoin transaction?Get transaction fees per transaction via gettransactionWhat would be the implications of limiting Bitcoin transactions to fifty inputs?Determine inputs/outputs of a transaction before sendingIs there a maximum fee per byte in satoshis?Why does Bitcoin need transaction fees prior to hitting its maximum coin limit?What happens if “transaction size” is larger than “maximum block size”?
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
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
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
asked Apr 2 at 4:19
Pedro GonçalvesPedro Gonçalves
646
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
transactions transaction-input input
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
asked Apr 2 at 4:19
Pedro GonçalvesPedro Gonçalves
646
asked Apr 2 at 4:19
Pedro GonçalvesPedro Gonçalves
646
646
add a comment |
add a comment |
1 Answer
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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.
answered Apr 2 at 5:19
Andrew Chow♦Andrew Chow
33.3k42462
2
To note, that includes zero security or signatures.
– Anonymous
2 days ago
add a comment |
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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.
answered Apr 2 at 5:19
Andrew Chow♦Andrew Chow
33.3k42462
2
To note, that includes zero security or signatures.
– Anonymous
2 days ago
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.
answered Apr 2 at 5:19
Andrew Chow♦Andrew Chow
33.3k42462
2
To note, that includes zero security or signatures.
– Anonymous
2 days ago
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.
answered Apr 2 at 5:19
Andrew Chow♦Andrew Chow
33.3k42462
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.
answered Apr 2 at 5:19
Andrew Chow♦Andrew Chow
33.3k42462
edited 2 days ago
answered Apr 2 at 5:19
Andrew Chow♦Andrew Chow
33.3k42462
answered Apr 2 at 5:19
Andrew Chow♦Andrew Chow
33.3k42462
answered Apr 2 at 5:19
Andrew Chow♦Andrew Chow
33.3k42462
33.3k42462
2
To note, that includes zero security or signatures.
– Anonymous
2 days ago
add a comment |
2
To note, that includes zero security or signatures.
– Anonymous
2 days ago
2
2
To note, that includes zero security or signatures.
– Anonymous
2 days ago
To note, that includes zero security or signatures.
– Anonymous
2 days ago
add a comment |
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