Q:

How do I print the percent sign(%) in c

I am a beginner in C, and I was wondering why this program does not print % sign?
The code is:
#include<stdio.h>

main()
{
     printf("%");
     getch();
}

A:

Your problem is that you have to change:   
printf("%"); 

to 
printf("%%");

Or you could use ASCII  code and write:
printf("%c", 37);

:)

A:

there's no explanation in this topic why to print a percentage sign one must type %% and not for example escape character with percentage - \%. 
from comp.lang.c FAQ list · Question 12.6 :

The reason it's tricky to print % signs with printf is that % is
  essentially printf's escape character. Whenever printf sees a %, it
  expects it to be followed by a character telling it what to do next.
  The two-character sequence %% is defined to print a single %.
To understand why \% can't work, remember that the backslash \ is the
  compiler's escape character, and controls how the compiler interprets
  source code characters at compile time. In this case, however, we want
  to control how printf interprets its format string at run-time. As far
  as the compiler is concerned, the escape sequence \% is undefined, and
  probably results in a single % character. It would be unlikely for
  both the \ and the % to make it through to printf, even if printf were
  prepared to treat the \ specially.

so the reason why one must type printf("%%"); to print single % is that's what is defined in printf function. % is an escape character of printf's, and \ of compiler.

A:

Use "%%".  The man page describes this requirement:

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

