Write a program to print a semicolon without using a semicolon anywhere in the code? | C Programming

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: Program to print a semicolon without using semicolon in the code.

#include<stdio.h>
int main(void) 

{
//prints the character with ascii value 59, i.e., semicolon
if (printf("%c\n", 59)) 

{
//prints semicolon
}
return 0;
}
Output:
;
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(;).