Question

Assume input is a char array holding a C-string. Write code that counts the number of elements in the array that contain an alphabetic character.

Short answer

Answer: __________

Step-by-step solution

Initialize variables

Declare and initialize a counter variable (let's call it 'count') and set it to 0. We will use this variable to keep track of the number of alphabetic characters in the C-string.

Traverse the char array

Use a 'for' loop to iterate through the elements in the char array holding the C-string until the null character ('\0') is reached, which signifies the end of the string.

Check if the character is alphabetic

Within the loop, check if the current character is an alphabetic character by using the 'isalpha()' function from the 'ctype.h' library. In C, 'isalpha()' returns a non-zero value (true) if the character is alphabetic and zero (false) if it's not.

Increment the counter for each alphabetic character

If 'isalpha()' returns true for the current character, increment the 'count' variable by 1.

Output the final count

Once the loop has reached the null character, print the value of the 'count' variable, which will show the total number of alphabetic characters in the C-string input. Here is the complete code: ```c #include #include int main() { char input[] = "Hello, World!"; // Replace this with the provided C-string int count = 0; for(int i = 0; input[i] != '\0'; i++) { if(isalpha(input[i])) { count++; } } printf("Number of alphabetic characters: %d\n", count); return 0; } ```