C Program to Print a Semicolon without using a Semicolon anywhere in the Code

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.

```
/*
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 * 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))
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```
    {
```
```
    }
```
```
    return 0;
```
```
}
```

Program Explanation

1. The ASCII value of semicolon is 59.
2. Print this ASCII value and exit.

Runtime Test Cases

Sanfoundry Global Education & Learning Series – 1000 C Programs.

Here’s the list of Best Reference Books in C Programming, Data-Structures and Algorithms

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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