What would happen if we provide to OP_CHECKMULTISIG more than threshold number (m) of signatures?

Question

Asked here but still no answer.

What would happen if we provide to OP_CHECKMULTISIG opcode more than threshold number (m) of signatures?

For example, if the m = 3 and n = 6 and the output script is:

3 <pubkey_1> ... <pubkey_6> 6 CHECKMULTISIG

And we provide, for example, this input script (or witness stack for segWit 0):

0 <signature_1> <signature_2> <signature_3> <signature_4> <signature_5>

Will it pass or fail?

Answer 1

Since the activation of BIP147 (nulldummy softfork) in 2017, the answer is no, this will fail. Before that point it would have been generally non-standard but allowed.

An input to OP_CHECKMULISIG expects on the stack: (from back to front):

• the number n
• n public keys
• the number k
• k signatures
• a dummy element (ignored pre-BIP147, must be 0 post-BIP147)

In your example, n=6 is read, followed by 6 keys, then k=3, and then 3 signatures. These 3 signatures do match the penultimate 3 keys, which on itself would work. However, after that a 0 is expected, yet instead there is another signature, which fails.

Before BIP147, the signature_2 would function as the dummy, and signature_1, plus a 1 (from the evaluation of OP_CHECKMULTISIG) and 0 would be left on the stack.