How to print % using printf()?

Here is the standard prototype of printf function in C:

int printf(const char *format, ...);

The format string is composed of zero or more directives: ordinary characters (not %), which are copied unchanged to the output stream; and conversion specifications, each of argument (and it is an error if insufficiently many arguments are given). 

The character % is followed by one of the following characters. 

• Flag character 
• Field width
• Precision
• Length modifier
• Conversion specifier

Check this for details of all the above characters. The main thing to note in the standard is the below line about conversion specifier.

A '%' is written. No argument is converted. The complete conversion specification is'%%'.

So we can print “%” using “%%” 

c

/* Program to print %*/
#include <stdio.h>
int main()
{
   printf("%%");
   getchar();
   return 0;
}

Output

%

We can also print “%” using below. 

printf(“%c”, ‘%’);
printf(“%s”, “%”);