c setvbuf() function call not working as expected

Question

I'm learning system programming and when I run the following code

#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>

int main()
{
    char buff[5];
    setvbuf(stderr, buff, _IOFBF, 5);
    for (int i = 1; i < 10; i++) {
        fprintf(stderr, "%d", i);
    }
    _exit(0);
}

setvbuf is imposing fully buffered policy on stderr with a 5-byte sized buffer, so I expected to get 12345 but instead it prints 1234567 , why is that?

Answer 1

strace-ing the program shows that when compiled against glibc it first does a write of size 1, then a write of size 5, and finally a write of size 1:

$ strace -o /dev/stdout -e write,writev ./a.out 2>/dev/null
write(2, "1", 1)                        = 1
write(2, "23456", 5)                    = 5
write(2, "7", 1)                        = 1

With musl it does 7 writes of size 1:

$ strace -o /dev/stdout -e write,writev ./a.out 2>/dev/null
writev(2, [{iov_base="1", iov_len=1}, {iov_base=NULL, iov_len=0}], 2) = 1
writev(2, [{iov_base="2", iov_len=1}, {iov_base=NULL, iov_len=0}], 2) = 1
writev(2, [{iov_base="3", iov_len=1}, {iov_base=NULL, iov_len=0}], 2) = 1
writev(2, [{iov_base="4", iov_len=1}, {iov_base=NULL, iov_len=0}], 2) = 1
writev(2, [{iov_base="5", iov_len=1}, {iov_base=NULL, iov_len=0}], 2) = 1
writev(2, [{iov_base="6", iov_len=1}, {iov_base=NULL, iov_len=0}], 2) = 1
writev(2, [{iov_base="7", iov_len=1}, {iov_base=NULL, iov_len=0}], 2) = 1
writev(2, [{iov_base="8", iov_len=1}, {iov_base=NULL, iov_len=0}], 2) = 1
writev(2, [{iov_base="9", iov_len=1}, {iov_base=NULL, iov_len=0}], 2) = 1
+++ exited with 0 +++

So it's obvious that both C library implementations treat the size argument of setvbuf(stream, buf, _IOFBF, size) as a maximum, and they feel free to flush the buffer even before it's full.

AFAICS there doesn't seem to be anything in the standard against this interpretation.