C Program to Print Semicolon without using Semicolon

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.

  1. /*
  2.  * C Program to Print a Semicolon without using a Semicolon
  3.  * anywhere in the code
  4.  */
  5. #include <stdio.h>
  6.  
  7. int main(void)
  8. {
  9.     //59 is the ascii value of semicolumn
  10.     if (printf("%c ", 59))
  11.     {
  12.     }
  13.     return 0;
  14. }
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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If you wish to look at other example programs on Simple C Programs, go to Simple C Programs. If you wish to look at programming examples on all topics, go to C Programming Examples.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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