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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?

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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 3 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.

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    To note, that includes zero security or signatures. – Anonymous Apr 2 at 6:11
  • and 2 bytes for input count correction: it is 3 bytes because 27022 inputs in CompactInt is equal to fd8e69 (it needs an initial 0xfd to indicate 2 bytes follow up) – Coding Enthusiast May 21 at 8:18
  • @Anonymous What do you mean by "includes zero security"? – Jonathan Cross May 21 at 16:26
  • Without signatures, all transaction in the block described here can be spent by anybody else. Including signatures reduces the number of inputs considerably. – Anonymous May 21 at 17:09

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