OP_MAX - Returning the Larger Value

Opcode:     OP_MAX
Hex:        0xA4
Decimal:    164
Input:      a b
Output:     max(a, b)

Input: [b=10, a=3]  → OP_MAX → [10]
Input: [b=3, a=10]  → OP_MAX → [10]
Input: [b=5, a=5]   → OP_MAX → [5]
Input: [b=-1, a=-5] → OP_MAX → [-1]

case OP_MAX:
{
    if (stack.size() < 2)
        return set_error(serror, SCRIPT_ERR_INVALID_STACK_OPERATION);
    CScriptNum bn1(stacktop(-2), fRequireMinimal);
    CScriptNum bn2(stacktop(-1), fRequireMinimal);
    popstack(stack);
    popstack(stack);
    stack.push_back((bn1 > bn2 ? bn1 : bn2).getvch());
}
break;

<user_value> <floor> OP_MAX
→ result is at least <floor>

<value> <floor> OP_MAX <ceiling> OP_MIN
→ result is clamped to [floor, ceiling]