C code to Count Alphabets, Numeric, Special character and Space

C code to Count Alphabets, Numeric, Special character and Space

In this article, we will discuss the concept of the C code to Count Alphabets, Numeric, Special character and Space in C language

In this post, we are going to learn how to count the total number of Alphabets, Numeric, Special character and Space of the given string in C programming language

C code to Count Alphabets, Numeric, Special character and Space

Count Alphabets, Numeric, Special character and Space using for loop

The program allows the user to enter a String and then it counts and display the total number of Alphabets, Numeric, Special character and Space of the given string using for loop in C programing language

Program 1

#include <stdio.h>
#include <stdlib.h>

int main()
{
    char str[100];
    int i;
    int Alphabets=0,num=0,special=0,space=0;
    printf("Please enter the string \n");
    gets(str);
    for(i=0; str[i] != '\0'; i++){
        
            if((str[i]>='A' && str[i]<='Z') || (str[i]>='a' && str[i]<='z')) {//Check Alphabets
                Alphabets++;
    }else if(str[i]>='0' && str[i]<='9') {//Check numbers
                num++;
    }

    else if(str[i]==' '){//Check space
                space++;
    }
    else{//Find special caracters
        special++;
    }
    }
    printf("\nAlphabets letters: %d",Alphabets);

    printf("\Numbers: %d",num);
      printf("\nSpace: %d",space);
    printf("\nSpecial characters: %d",special);
    getch();
    return 0;
}

When the above code is executed, it produces the following result

Please enter the string
code4coding.com 2020

Alphabets letters:13
Numbers:5
Space:1
Special characters:1

Approach

1. Declare a Character array as char str[100];
2. Declare and initialize integer variables as Alphabets=0,num=0,space=0,Special=0;
   
3. The user asked to enter a string
4. The given string is stored in the variable str;
5. A for loop is used to count every total of the given string.
6. It is initialized as i=0, checks the condition whether str[i] != ‘\0’; and executes the loop until the given condition becomes true
7. Use an if condition to test if((str[i]>=’A’ && str[i]<=’Z’) || (str[i]>=’a’ && str[i]<=’z’)), if it is true,  The Alphabets becomes Alphabets + 1( Alphabets = Alphabets +1);
8. When it is false, control moves to the else-if part and evaluates the test-expression of else-if. if it is true The num becomes num+1(num=num+1);
9. When it is false, control moves to the next else-if part and evaluates the test-expression of else-if. if it is true The space becomes space+1(space=space+1);
10. When it is false, control moves to the else part and executes the else part statements. the special becomes special+1
11. Finally, the program displays the total number of the Alphabets, Numeric, Special character and Space of the given string

Count Alphabets, Numeric, Special character and Space using while loop

The program allows the user to enter a String and then it counts and display the total number of Alphabets, Numeric, Special character and Space of the given string using While loop in C programing language

Program 2

#include <stdio.h>
#include <stdlib.h>

int main()
{
    char str[100];
    int i;
    int Alphabets=0,num=0,special=0,space=0;
    printf("Please enter the string \n");
    gets(str);
    i=0;
    while(str[i] != '\0'){
        //check for Alpabets
            if((str[i]>='A' && str[i]<='Z') || (str[i]>='a' && str[i]<='z')) {
                Alphabets++;
    }else if(str[i]>='0' && str[i]<='9') {//Check numbers
                num++;
    }

    else if(str[i]==' '){//Check spave
                space++;
    }
    else{
        special++;//Check special characters
    }
    i++;
    }
    printf("\nAlphabets letters: %d",Alphabets);

    printf("\nnumbers: %d",num);
      printf("\nSpace: %d",space);
    printf("\nSpecial characters: %d",special);
    getch();
    return 0;
}

When the above code is executed, it produces the following result

Please enter the string
Code For You:code4coding.com

Alphabets letters:23
Numbers:1
Space:2
Special characters:2

Approach

1. Declare a Character array as char str[100];
2. Declare and initialize integer variables as Alphabets=0,num=0,space=0,Special=0;
   
3. The user asked to enter a string
4. The given string is stored in the variable str;
5. A while loop is used to count total characters of the given string.
6. It is initialized as i=0, checks the condition whether str[i] != ‘\0’; and executes the loop until the given condition becomes true
7. Use an if condition to test if((str[i]>=’A’ && str[i]<=’Z’) || (str[i]>=’a’ && str[i]<=’z’)), if it is true,  The Alphabets becomes Alphabets + 1( Alphabets = Alphabets +1);
8. When it is false, control moves to the else-if part and evaluates the test-expression of else-if. if it is true The num becomes num+1(num=num+1);
9. When it is false, control moves to the next else-if part and evaluates the test-expression of else-if. if it is true The space becomes space+1(space=space+1);
10. When it is false, control moves to the else part and executes the else part statements. the special becomes special+1
11. Finally, the program displays the total number of the Alphabets, Numeric, Special character and Space of the given string

Count Alphabets, Numeric, Special character and Space using do-while loop

The program allows the user to enter a String and then it counts and display the total number of Alphabets, Numeric, Special character and Space of the given string using do-while loop in C programing language

Program 3

#include <stdio.h>
#include <stdlib.h>

int main()
{
    char str[100];
    int i;
    int Alphabets=0,num=0,special=0,space=0;
    printf("Please enter the string \n");
    gets(str);
    i=0;
    do{
        //check for uppercase
            if((str[i]>='A' && str[i]<='Z') || (str[i]>='a' && str[i]<='z')) {//Check Alphabets
                Alphabets++;
    }else if(str[i]>='0' && str[i]<='9') {//Check numeric
                num++;
    }

    else if(str[i]==' '){//Check space
                space++;
    }
    else{//Check special
        special++;
    }
    i++;
    }while(str[i] != '\0');
    printf("\nAlphabets letters: %d",Alphabets);

    printf("\nnumbers: %d",num);
      printf("\nSpace: %d",space);
    printf("\nSpecial characters: %d",special);
    getch();
    return 0;
}

When the above code is executed, it produces the following result

Please enter the string
Code with me:code4coding.com

Alphabets letters:23
Numbers:1
Space:2
Special characters:2

Approach

1. Declare a Character array as char str[100];
2. Declare and initialize integer variables as Alphabets=0,num=0,space=0,Special=0;
   
3. The user asked to enter a string
4. The given string is stored in the variable str;
5. A do-while loop is used to count total characters of the given string.
6. It is initialized as i=0, checks the condition whether str[i] != ‘\0’; and executes the loop until the given condition becomes true
7. Use an if condition to test if((str[i]>=’A’ && str[i]<=’Z’) || (str[i]>=’a’ && str[i]<=’z’)), if it is true,  The Alphabets becomes Alphabets + 1( Alphabets = Alphabets +1);
8. When it is false, control moves to the else-if part and evaluates the test-expression of else-if. if it is true The num becomes num+1(num=num+1);
9. When it is false, control moves to the next else-if part and evaluates the test-expression of else-if. if it is true The space becomes space+1(space=space+1);
10. When it is false, control moves to the else part and executes the else part statements. the special becomes special+1
11. Finally, the program displays the total number of the Alphabets, Numeric, Special character and Space of the given string

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