What does a double-percent sign (%%) do in gcc inl

Question

I came across a code that looks like this:

asm volatile (
    # [...]
    "movl $1200, %%ecx;"
    # [...]
);

I know what movl $1200, %ecx does in x86. but I was confused by why there are two percent signs.

Answer 1

It depends

• if there is a colon : after the string, then it is an extended asm, and %% escapes the percent which could have especial meanings as mentioned by Carl. Example:

  uint32_t in = 1;
  uint32_t out = 0;
  asm volatile (
      "movl %1, %%eax;"
      "inc %%eax;"
      "movl %%eax, %0"
      : "=m" (out) /* Outputs. '=' means written to. */
      : "m" (in)   /* Inputs. No '='. */
      : "%eax"
  );
  assert(out == in + 1);
  

• otherwise, it will be a compile time error, because without colon it is a basic asm which does not support variable constraints and does not need or support escaping %1. E.g.:

  asm volatile ("movl $1200, %ecx;");
  

  works just fine.

Extended asm is more often used since it is much more powerful.

Answer 2

This helps GCC to distinguish between the operands and registers. operands have a single % as prefix. '%%' is always used with registers.

Answer 3

GCC inline assembly uses %0, %1, %2, etc. to refer to input and output operands. That means you need to use two %% for real registers.

Check this howto for great information.