C Program to Print a Semicolon Without Using a Semicolon

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Generally when use printf() statement we have to use semicolon at the end.

If we want to print a semicolon, we use the statement: printf(";");

In above statement, we are using two semicolons. The task of printing a semicolon without using semicolon anywhere in the code can be accomplished by using the ascii value of  ; which is equal to 59.

Program:
#include <stdio.h>

int main(void) {
    //prints the character with ascii value 59, i.e., semicolon
    if (printf("%c ", 59)) {
      //prints semicolon
    }
    return 0;
}
Output of program:
;
Explanation:

If statement checks whether return value of printf function is greater than zero or not. The return value of function call printf("%c",59) is 1. As printf returns the length of the string printed. printf("%c",59) prints ascii value that corresponds to 59, that is semicolon ;.


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