Given a character, we have to check whether the given character is a whitespace character or not without using the library function.

Program:

The source code to check a given character is a whitespace character or not 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 check a given character is a whitespace character or not
// without using library function

#include <stdio.h>

int isWhiteSpace(char ch)
{
    if ((ch == ' ') || (ch == '\t') || (ch == '\n') || (ch == 'v') || (ch == '\r') || (ch == '\f'))
        return 1;

    return 0;
}

int main()
{
    char ch1 = ' ';
    char ch2 = '\t';
    char ch3 = 't';

    if (isWhiteSpace(ch1))
        printf("Given character is a whitespace character\n");
    else
        printf("Given character is not a whitespace character\n");

    if (isWhiteSpace(ch2))
        printf("Given character is a whitespace character\n");
    else
        printf("Given character is not a whitespace character\n");

    if (isWhiteSpace(ch3))
        printf("Given character is a whitespace character\n");
    else
        printf("Given character is not a whitespace character\n");

    return 0;
}

Output:

Given character is a whitespace character
Given character is a whitespace character
Given character is not a whitespace character

Explanation:

In the above program, we created two functions isWhiteSpace() and main(). The isWhiteSpace() function is used to check the given character is a whitespace character or not.

In the main() function, we created three character variables ch1, ch2, ch3 that are initialized with ' ', '\t', 't' respectively. Then we checked given characters are whitespace characters or not by calling the isWhiteSpace() function and print the appropriate message on the console screen.