Immitating printf

Question

What's wrong with the follwing code? How can we make the function print() to work as printf?

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

void print(char *format,...)
{
    va_list args;
    va_start(args,format);
    printf(format,args);
}

int main() {
   print("%d %s",5,"le");
}

possible duplicate of [what's the difference between the printf and vprintf function families, and when should I use one over the other?](http://stackoverflow.com/questions/1485805/whats-the-difference-between-the-printf-and-vprintf-function-families-and-when) — Praetorian, May 21 '12 at 16:07
possible duplicate of [call printf using va_list](http://stackoverflow.com/questions/5977326/call-printf-using-va-list) — Cody Gray - on strike, Aug 05 '12 at 05:55

score 7 · Answer 1 · answered May 21 '12 at 15:58

7

If you need to pass varargs, then use vprintf instead.

answered May 21 '12 at 15:58

Oliver Charlesworth

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score 3 · Answer 2 · answered May 21 '12 at 15:58

3

You probably need to look at vprintf. That function (and the related ones) allow you to pass along a variable argument list, and they handle the formatting.

answered May 21 '12 at 15:58

Mark Wilkins

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score 3 · Accepted Answer · edited May 23 '17 at 12:27

3

You need vprintf there. Take a look this question, it has a similiar problem: what's the difference between the printf and vprintf function families, and when should I use one over the other?

edited May 23 '17 at 12:27

Community

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1

answered May 21 '12 at 15:59

kratenko

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score 1 · Answer 4 · answered May 21 '12 at 16:03

First of all there is a va_end() call missing which is mandatory if using va_start().

And you cannot use printf() if you want to use a va_list as argument. Take a look at vprintf().

example:

void print(char *format,...)
{
    va_list args;
    va_start(args,format);
    vprintf(format,args);
    va_end(args);
}

Immitating printf

4 Answers4