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Home/ Questions/Q 7987513
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T12:10:16+00:00 2026-06-04T12:10:16+00:00

I’m trying to wrap printf in a C program (well, actually _snprintf but this

  • 0

I’m trying to wrap printf in a C program (well, actually _snprintf but this example is simpler) and am having trouble getting the variable argument stuff to work. Here is my code:

#include <stdio.h>
#include <stdarg.h>

void works(void)
{
    printf("%d\n", 100);
}

void wrap_printf(const char *fmt, ...)
{
    va_list args;
    va_start(args, fmt);
    printf(fmt, args);
    va_end(args);
}

void broken(void)
{
    wrap_printf("%d\n", 100);
}

int main(void)
{
    works();
    broken();
    return 0;
}

Here is my output:

100
3668388

The args variable looks good after the call to va_start in my code, but as soon as I step into the C runtime code and they call va_start the value looks bad. Any thoughts as to what I might be doing wrong?

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  1. Editorial Team
    Editorial Team
    2026-06-04T12:10:18+00:00Added an answer on June 4, 2026 at 12:10 pm
    va_start(args, fmt);
    vprintf(fmt, args);
    va_end(args);
    

    You need to call vprintf instead of printf. The v*printf functions understand va_List arguments. I’m surprised you didn’t get a warning.

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