C Library - ispunct() function

The C ctype library ispunct() function is used to determine whether a given character is a punctuation character.

Punctuation characters are those that are not alphanumeric and are not whitespace characters. This typically includes characters such as !, @, #, $, %, ^, &, *, (, ), etc.

Syntax

Following is the C library syntax of the ispunct() function −

int ispunct(int ch);

Parameters

This function accepts a single parameter −

• ch − This is the character to be checked, passed as an int. It must be representable as an unsigned char or the value of EOF.

Return Value

The function returns a non-zero value (true) if the character is a punctuation character. Otherwise it returns zero.

Example 1: Checking Multiple Characters in a String

Here we, count the number of punctuation characters in the string "Hello, World!", which are , and !.

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

int main() {
   char str[] = "Hello, World!";
   int count = 0;

   for (int i = 0; str[i] != '\0'; i++) {
      if (ispunct(str[i])) {
         count++;
      }
   }
   printf("The string \"%s\" contains %d punctuation characters.\n", str, count);
   return 0;
}

Output

The above code produces following result −

The string "Hello, World!" contains 2 punctuation characters.

Example 2: User Input Validation

Now in this code, we check whether a user-input character is a punctuation character.

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

int main() {
   char ch;
   printf("Enter a character: ");
   scanf("%c", &ch);

   if (ispunct(ch)) {
      printf("You entered a punctuation character.\n");
   } else {
      printf("You did not enter a punctuation character.\n");
   }
   return 0;
}

Output

After execution of above code, we get the following result −

(Example input: @)
You entered a punctuation character.