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Asked: 2026-05-11T08:26:46+00:00

A previous question showed a nice way of printing to a string. The answer

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A previous question showed a nice way of printing to a string. The answer involved va_copy:

std::string format (const char *fmt, ...); {    va_list ap;    va_start (ap, fmt);    std::string buf = vformat (fmt, ap);    va_end (ap);    return buf; }   std::string vformat (const char *fmt, va_list ap) {    // Allocate a buffer on the stack that's big enough for us almost    // all the time.    s ize_t size = 1024;    char buf[size];     // Try to vsnprintf into our buffer.    va_list apcopy;    va_copy (apcopy, ap);    int needed = vsnprintf (&buf[0], size, fmt, ap);     if (needed <= size) {        // It fit fine the first time, we're done.        return std::string (&buf[0]);    } else {        // vsnprintf reported that it wanted to write more characters        // than we allotted.  So do a malloc of the right size and try again.        // This doesn't happen very often if we chose our initial size        // well.        std::vector <char> buf;        size = needed;        buf.resize (size);        needed = vsnprintf (&buf[0], size, fmt, apcopy);        return std::string (&buf[0]);    }

}

The problem I’m having is that the above code doesn’t port to Visual C++ because it doesn’t provide va_copy (or even __va_copy). So, does anyone know how to safely port the above code? Presumably, I need to do a va_copy copy because vsnprintf destructively modifies the passed va_list.

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1 Answer

  1. You should be able to get away with just doing a regular assignment:

    va_list apcopy = ap;

    It’s technically non-portable and undefined behavior, but it will work with most compilers and architectures. In the x86 calling convention, va_lists are just pointers into the stack and are safe to copy.

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