Program to check if a string contains any special character in C

Given a string str[], the task is to check whether the string contains any special character and if the string have a special character then print “The String is not accepted” else print “The string is accepted”.

Special characters are those characters which are neither numeric nor alphabetic i.e. − !@#$%^&*()+=-\][‘;/.,{}|:”<>?`~

So in C Programming language we will use if-else approach to solve the problem.

Input − str[] = {“tutorials-point”}

Output − the string is not accepted

Input − str[] = {“tutorialspoint”}

Output − The string is accepted

Approach used below is as follows to solve the problem −

• Traverse the whole string.

• Will look for the special character, if the special character does exist in the string then, print “The string is not accepted and break”. Else, print the string is accepted.

Other approach

If we are coding in java or any other language which supports the concept of regular expressions, then instead of if-else approach we will use regular expressions to check that whether they are present in the given string or not. This is not only a simple approach but also a fast one.

Algorithm

Start
In function int special_character(char str[], int n)
   Step 1→ initialize i and flag and set flag as 0
   Step 2→ Loop For i = 0 and i < n and ++i
      If(str[i] == '!' || str[i] == '@' || str[i] == '#' || str[i] == '$'
      || str[i] == '%' || str[i] == '^' || str[i] == '&' || str[i] == '*'
      || str[i] == '(' || str[i] == ')' || str[i] == '-' || str[i] == '{'
      || str[i] == '}' || str[i] == '[' || str[i] == ']' || str[i] == ':'
      || str[i] == ';' || str[i] == '"' || str[i] == '\'' || str[i] == '<'
      || str[i] == '>' || str[i] == '.' || str[i] == '/' || str[i] == '?'
      || str[i] == '~' || str[i] == '`' then
         Print "String is not allowed”
            Set flag as 1
         break
   Step 3→ If flag == 0 then,
      Print "string is accepted”
In function int main(int argc, char const *argv[])
   Step 1→ Declare and set str[] as {"Tutorials-point"}
   Step 2→ set n as strlen(str)
   Step 3→ special_character(str, n)
Stop

Example

 Live Demo

#include <stdio.h>
#include <string.h>
int special_character(char str[], int n){
   int i, flag = 0;
   for (i = 0; i < n; ++i){
      //checking each character of the string for special character.
      if(str[i] == '!' || str[i] == '@' || str[i] == '#' || str[i] == '$'
      || str[i] == '%' || str[i] == '^' || str[i] == '&' || str[i] == '*'
      || str[i] == '(' || str[i] == ')' || str[i] == '-' || str[i] == '{'
      || str[i] == '}' || str[i] == '[' || str[i] == ']' || str[i] == ':'
      || str[i] == ';' || str[i] == '"' || str[i] == '\'' || str[i] == '<'
      || str[i] == '>' || str[i] == '.' || str[i] == '/' || str[i] == '?'
      || str[i] == '~' || str[i] == '`' ){
         printf("String is not allowed\n");
         flag = 1;
         break;
      }
   }
   //if there is no special charcter
   if (flag == 0){
      printf("string is accepted\n");
   }
   return 0;
}
int main(int argc, char const *argv[]){
   char str[] = {"Tutorials-point"};
   int n = strlen(str);
   special_character(str, n);
   return 0;
}

Output

If run the above code it will generate the following output −

String is not allowed

Sunidhi Bansal

Published on 13-Aug-2020 11:39:56