Compiling old C code Y2038 conform still results in 4 byte variables

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According to this overview in order to compile Y2038 conform old code, we just need to add the preprocessor macro __USE_TIME_BITS64 to gcc, but that does not seem to work on an ARMv7 board with Debian 12 (bookworm):

#include <sys/types.h>
#include <sys/stat.h>

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

int main(void)
{
    struct stat sb;

    printf("sizeof time_t: %zu\n", sizeof(time_t));
    printf("sizeof stat timestamp: %zu\n", sizeof(sb.st_atime));

    return 0;
}

time_t is still 4 bytes:

root@debian:~# gcc -D__USE_TIME_BITS64 time.c -o time
root@debian:~# ./time
sizeof time_t: 4
sizeof stat timestamp: 4
root@debian:~#

glibc is 2.33, what am I doing wrong here?

libc glibc linux c
1 Answers

According to this post (which is getting a little old now, and some parts of which are probably no longer relevant):

... defining _TIME_BITS=64 would cause all time functions to use 64-bit times by default. The _TIME_BITS=64 option is implemented by transparently mapping the standard functions and types to their internal 64-bit variants. Glibc would also set __USE_TIME_BITS64, which user code can test for to determine if the 64-bit variants are available.

Presumably, this includes making time_t 64 bit.

So if your version of glibc supports this at all, it looks like you're setting the wrong macro. You want:

-D_TIME_BITS=64

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