C - Print Character '%'

Question

I was wondering if I could print % in c but when I use printf("%") It just shows this syntax in output.

    #include<stdio.h>
    int main()
   {
       printf("%");
   }

but the result is like this:

main.c:13:13: warning: spurious trailing ‘%’ in format [-Wformat=]

Does anybody know how I can literally print % as a character in c.

Have you tried looking at https://stackoverflow.com/questions/20274423/printing-a-sign-in-c-using-printf? — Justin, Nov 25 '21 at 06:43
Does this answer your question? [Printing a % sign in C using 'printf'](https://stackoverflow.com/questions/20274423/printing-a-sign-in-c-using-printf) — justANewb stands with Ukraine, Nov 25 '21 at 06:45

score 1 · Answer 1 · answered Nov 25 '21 at 06:43

1

You can do:

printf("%%");

or because % is a char, so you can use %c:

printf("%c", '%');

Overall, you have two ways.

#include<stdio.h>
int main()
{
   printf("%%\n");
   printf("%c\n", '%');
}

will prints:

%
%

answered Nov 25 '21 at 06:43

score 1 · Accepted Answer · edited Nov 25 '21 at 07:58

1

To print a % character you need to follow this structure:

int main()
{
    printf("%%");
    return 0;
}

you have to use the character twice.

edited Nov 25 '21 at 07:58

Wai Ha Lee

answered Nov 25 '21 at 06:43

score 0 · Answer 3 · answered Nov 25 '21 at 06:43

0

The good old method would be to use the %c format specifier in printf() function

    #include<stdio.h>
    int main()
   {
       printf("%c",'%');
   }

answered Nov 25 '21 at 06:43

Abhijit

score 0 · Answer 4 · answered Nov 25 '21 at 06:44

0

#include <stdio.h>

int main()
{
    printf("%%");
    return 0;
}

answered Nov 25 '21 at 06:44

Agula Otgonbaatar

4 Answers4