_IO_puts and Dynamic Instruction count

Question

I am new to assembly and have a simple C program printing a string to stdout. In assembly, this translates to main calling _IO_Puts. The implementation of _IO_puts is as below taken from: https://code.woboq.org/userspace/glibc/libio/ioputs.c.html is given below.

int
_IO_puts (const char *str)
{
  int result = EOF;
  size_t len = strlen (str);
  _IO_acquire_lock (stdout);
  if ((_IO_vtable_offset (stdout) != 0
       || _IO_fwide (stdout, -1) == -1)
      && _IO_sputn (stdout, str, len) == len
      && _IO_putc_unlocked ('\n', stdout) != EOF)
    result = MIN (INT_MAX, len + 1);
  _IO_release_lock (stdout);
  return result;
}

I am unable to figure out why the number of dynamic instructions change and sometimes decrease with increase in the string length on a simulates MIPS processor?

You need to dig deeper, presumably into `_IO_sputn`. It may be looping through your string. `stdout` is line buffered, so it first stores your string and only outputs it when the `\n` is added by the `putc`. That involves a `memcpy` which scales with the length of your string. — Jester, Sep 02 '21 at 12:05
glibc strlen on MIPS uses the pure C bithack that checks 4 bytes at a time. (Unlike most other ISAs that have hand-written asm). [Why does glibc's strlen need to be so complicated to run quickly?](https://stackoverflow.com/q/57650895). It's pretty non-simple, but its startup strategy depends on alignment. — Peter Cordes, Sep 02 '21 at 18:37

score 3 · Accepted Answer · answered Sep 02 '21 at 18:40

As noted in comments, glibc _IO_sputn involves doing strlen, and memcpy of that size.

glibc strlen on MIPS uses the pure C bithack that checks 4 bytes at a time. (Unlike most other ISAs that have hand-written asm). Why does glibc's strlen need to be so complicated to run quickly?. It's pretty non-simple, and its startup strategy depends on alignment of the start of the string.

More relevant here, finding the terminating 0 byte at the 4th byte of one word vs. the 1st byte of the next probably also costs fewer instructions. (Or 8th vs. 1st for MIPS64). So that's probably why you're seeing non-monotonic scaling of the dynamic instruction count.

memcpy will also take fewer instructions for a multiple of 4 (or 8) bytes, and it picks strategies via branching on the size and maybe alignment of src and dst. (MIPS before MIPS32/64r6 doesn't have guaranteed efficient unaligned stores). So sequential calls to stdio functions that don't involve flushing the buffer can lead to different destination alignments of the stdout buffer.

_IO_puts and Dynamic Instruction count

1 Answers1