C Programming: Count the number of punctuation characters exists in a string

Last update on December 20 2024 13:21:36 (UTC/GMT +8 hours)

Write a program in C to count the number of punctuation characters present in a string.

Sample Solution:

C Code:

#include<stdio.h>
#include<ctype.h>

int main() {
    int ctr1 = 0; // Counter for iterating through the string
    int ctr2 = 0; // Counter for counting punctuation characters
    char str[100]; // Array to store the input string

    printf("\n Count the number of punctuation characters exists in a string :\n");
    printf("------------------------------------------------------------------\n");
    printf(" Input a string : ");
    fgets(str, sizeof str, stdin); // Read a string from the user

    // Loop through the string and count punctuation characters
    while (str[ctr1]) {
        if (ispunct(str[ctr1])) {
            ctr2++; // Increment the counter if the character is a punctuation
        }
        ctr1++; // Move to the next character in the string
    }

    printf(" The punctuation characters exists in the string is : %d\n\n", ctr2); // Display the count of punctuation characters
    return 0; // Return 0 to indicate successful execution of the program
}

Sample Output:

 Count the number of punctuation characters exists in a string :
------------------------------------------------------------------
 Input a string : The quick brown fox,jumps over the,lazy dog.
 The punctuation characters exists in the string is : 3

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