This is a C Program to print a semicolon without using a semicolon anywhere in the code.
Problem Description
This program prints a semicolon without using a semicolon anywhere in the code.
Problem Solution
1. Print the ASCII value of semicolon and exit.
Program/Source Code
Here is source code of the C Program to print a semicolon without using a semicolon anywhere in the code. The C program is successfully compiled and run on a Linux system. The program output is also shown below.
/*
* C Program to Print a Semicolon without using a Semicolon
* anywhere in the code
*/
#include <stdio.h>
int main(void)
{
//59 is the ascii value of semicolumn
if (printf("%c ", 59))
{
}
return 0;
}
Program Explanation
1. The ASCII value of semicolon is 59.
2. Print this ASCII value and exit.
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Runtime Test Cases
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