Question

4. [8] Write a C function with prototype · void letter_freq(const char word[], int freq []); This function computes the number of appearances of each letter in the string word and stores them in array freq of size 26. The letters are the 26 letters of the Latin alphabet whose ASCII values are in the range 97-122 for the lower case letters, and in the range 65-90 for the uppercase letters. You must account for uppercase and lowercase letter variants, which should be counted together. The counts have to be stored in array freq in alphabetical order of letters, which corresponds to the increasing order of their ASCII values. Specifically, freq[O) should store the count of 'A' and 'a', freq[1] should store the count of 'B' and 'b', and so on. This function has also to print the counts indicating each letter and its count on a separate line, as follows: The count of 'A' and 'a' is ... The count of 'B' and 'b'is ... Write a program to test the function. Hint: If variable x of type char represents a lower case letter, then the corresponding index in the array equals the integer value of x-'a'. If x is an upper case letter, then the index in the array equals x-'A'. could you write it in C programming? thank you!

u also need to store the letters' ASCII number in the freq array

Answer 1

Here Iam providing the code and the output to the given problem.

Code:-

#include <stdio.h> #include<string . h> #include<ctype.h> void letter_freq(char word[], char UpperCase[] , char LowerCase [ ] ); //function declaration int main() char word[100] , UpperCase[26] , LowerCase[26]; //declaring arrays printf("Enter the word :"); scanf ("%s" , word); //storing the input int i; for (i e ; i<- 26 ; i++){ //taking initial values of arrays as 0 Upper Case[i] -0; Lower Case[i] -0; letter_freq(word ,UpperCase , LowerCase); //function calling void letter_freq (char word[], char UpperCase[] , char LowerCase[] ){ int i, word_lenght, index; word_lenght strlen(word);//calculating word length fori -e; i word_lenght ; it+){ //Looping through the word length if (word[i] 65 && word[i] <= 90 ){ // if the letter is Capital 'A'; //taking the index word[i] index UpperCase[index] = UpperCase[index]+1; //counting the repeated times and storing else{//if letter is LowerCase index 'a'; word[i] LowerCase [index] = LowerCase[index]+1; //counting the repeated times and storing for(i = 0; i<26 ; i++){ printf("The count of %c and %c is %d and %d\n i+65, i+97 ,UpperCase [i] , LowerCase [i]);//printing the result

Code in text format:-

#include <stdio.h>
#include<string.h>
#include<ctype.h>
void letter_freq(char word[] , char UpperCase[] , char LowerCase[]); //function declaration

int main()
{
char word[100] , UpperCase[26] , LowerCase[26]; //declaring arrays
printf("Enter the word :");
scanf("%s" , word); //storing the input
int i;
for ( i = 0 ; i<= 26 ; i++){ //taking initial values of arrays as 0
UpperCase[i] =0;
LowerCase[i] =0;
}
  
letter_freq(word ,UpperCase ,LowerCase); //function calling

}

void letter_freq(char word[] , char UpperCase[] , char LowerCase[]){
  
int i , word_lenght,index;
word_lenght = strlen(word);//calculating word length
  
for( i =0 ; i < word_lenght ; i++){ //looping through the word length
if(word[i] >= 65 && word[i] <= 90 ){ // if the letter is Capital
index = word[i] - 'A'; //taking the index
UpperCase[index] = UpperCase[index]+1; //counting the repeated times and storing
}
else{ //if letter is LowerCase
  
index = word[i] - 'a';
LowerCase[index] = LowerCase[index]+1; //counting the repeated times and storing
}
  
}
for(i = 0 ; i<26 ; i++){
printf("The count of %c and %c is %d and %d\n " , i+65 , i+97 ,UpperCase[i] , LowerCase[i]);//printing the result
}
  
  
  
  
  
}

  
  
Output:-

Enter the word :ABCDab
The count of A and a is 1 and 1
The count of B and b is 1 and 1
The count of C and c is 1 and 0
The count of D and d is 1 and 0
The count of E and e is 0 and 0
The count of F and f is 0 and 0
The count of G and g is 0 and 0
The count of H and h is 0 and 0
The count of I and i is 0 and 0
The count of J and j is 0 and 0
The count of K and k is 0 and 0
The count of L and l is 0 and 0
The count of M and m is 0 and 0
The count of N and n is 0 and 0
The count of O and o is 0 and 0
The count of P and p is 0 and 0
The count of Q and q is 0 and 0
The count of R and r is 0 and 0
The count of S and s is 0 and 0
The count of T and t is 0 and 0
The count of U and u is 0 and 0
The count of V and v is 0 and 0
The count of W and w is 0 and 0
The count of X and x is 0 and 0
The count of Y and y is 0 and 0
The count of Z and z is 0 and 0