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C program to print all punctuation marks without using library function

Here, we are going to learn how to print all punctuation marks without using library function in C language?
Submitted by Nidhi, on August 25, 2021

Problem statement

Here, we will print all punctuation marks are available in C language without using any library function.

Program

The source code to print all punctuation marks without using the library function is given below. The given program is compiled and executed using GCC compile on UBUNTU 18.04 OS successfully.

// C program to print all punctuation marks
// without using library function

#include <stdio.h>

int isPunctuation(char ch)
{
    if (ch == '!' || ch == '\"' || ch == '#' || ch == '$' || ch == '%' || ch == '&' || ch == '\'' || ch == '(' || ch == ')' || ch == '*' || ch == '+' || ch == ',' || ch == '-' || ch == '.' || ch == '/' || ch == ':' || ch == ';' || ch == '<' || ch == '=' || ch == '>' || ch == '?' || ch == '@' || ch == '[' || ch == '\\' || ch == ']' || ch == '^' || ch == '`' || ch == '{' || ch == '|' || ch == '}' || ch == '~')
        return 1;

    return 0;
}

int main()
{
    int cnt;
    printf("Punctuation marks are: \n");

    for (cnt = 0; cnt <= 127; cnt++)
        if (isPunctuation(cnt) != 0)
            printf("%c ", cnt);

    printf("\n");

    return 0;
}

Output

Punctuation marks are: 
! " # $ % & ' ( ) * + , - . / : ; < = > ? @ [ \ ] ^ ` { | } ~

Explanation

Here, we created two functions isPunctuation() and main(). The isPunctuation() function is used to check the given character is a punctuation mark or not.

In the main() function, we printed all punctuation marks available in C language without using the library function.

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