Using wsprintf to convert int to wchar_t*

Theodoros Chatzigiannakis picture Theodoros Chatzigiannakis · Jan 5, 2013 · Viewed 18.1k times · Source

I'm trying to get a wchar_t* formatted with an int as a parameter. I've Googled a lot but I've only ended up more confused. So, consider this code:

int main(int argc, char** argv) {

   wchar_t buf[16];
   wsprintf(buf, L"%d", 5);
   wprintf(L"[%ls]\n", buf);

   system("pause");
   return 0;

};

Having assumed that wchar_t, wsprintf and wprintf are the wide character equivalents of char, sprintf and printf respectively, I expected the above to print [5], but it prints garbage between [ and ]. What is the correct way to achieve the desired result? And what am I misunderstanding here?

(I should clarify that portability is more important than security here, so I'd like to know a solution that uses this family of functions instead of safer vendor-specific extensions.)

Answer

user529758 picture user529758 · Jan 5, 2013

wsprintf() is a Windows-specific function, it's unavailable on Unixes. What you want to achieve can be done in a more portable way (I have tried this slightly modified code snippet and it worked as expected):

#include <wchar.h>
#include <stdio.h>

int main(int argc, char **argv)
{
    wchar_t buf[16];
    swprintf(buf, sizeof(buf) / sizeof(*buf), L"%d", 5);
    wprintf(L"[%ls]\n", buf);

    return 0;
}

Output:

[5]