I completed a round trip transaction with the redeem script hash: 31vFvEkyAb2iusbPip36H9rVt6pxExuaVR, whose redeem script is OP_3 (0x53) (yes I know not secure).

I can understand OP_1 redeem script unlocking its P2SH UTXO as 1 is left on the stack after redeem script execution, but with OP_3, 3 is left on the stack. So how does this successfully execute?

Tx: 36ca5aede94112009621bda42b3e84f49f7aa45886837d1d51c6590a04fcb8e7


1 Answer 1


A script execution is successful if it ends with a non-empty stack whose top element is not 0. Therefore, a script consisting of only OP_3 will be successful.

You may also be interested in the topic “For a redeem script to satisfy the conditions of a script what must it leave on the stack post execution?” which describes the criteria for successful scripts in more detail (H/T Michael Folkson). I would recommend using testnet or signet for experimental transactions.

  • Yes this is very helpful, thanks. However a P2SH redeem script of say, 3 OP_3s (535353) fails with error "Stack size must be exactly one after execution" because it is valid, but non-standard, and hence would have to be submitted to miner directly. Is that correct?
    – Peter
    Feb 8, 2023 at 23:36
  • 1
    Yes, stacks with more than one remaining element are non-standard for P2SH, and consensus invalid for segwit outputs.
    – Murch
    Feb 9, 2023 at 2:03

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