Counting Alphabetic Characters That Are Contained in an Array with C

Question

Hello everyone,

I am having trouble with a homework question that I've been working at for quite some time. I don't know exactly why the question is asking and need some clarification on that and also a push in the right direction.

Here is the question:

(2) Solve this problem using one single subscripted array of counters. The program uses an array of characters defined using the C initialization feature. The program counts the number of each of the alphabetic characters a to z (only lower case characters are counted) and prints a report (in a neat table) of the number of occurrences of each lower case character found. Only print the counts for the letters that occur at least once. That is do not print a count if it is zero. DO NOT use a switch statement in your solution. NOTE: if x is of type char, x-‘a’ is the difference between the ASCII codes for the character in x and the character ‘a’. For example if x holds the character ‘c’ then x-‘a’ has the value 2, while if x holds the character ‘d’, then x-‘a’ has the value 3. Provide test results using the following string:

“This is an example of text for exercise (2).”

And here is my source code so far:

#include<stdio.h>

int main() {

    char c[] = "This is an example of text for exercise (2).";
    char d[26];

    int i;
    int j = 0;
    int k;

    j = 0;

    //char s = 97;

    for(i = 0; i < sizeof(c); i++) {
        for(s = 'a'; s < 'z'; s++){
            if( c[i] == s){

                k++;
                printf("%c,%d\n", s, k);
                k = 0;

            }
        }
    }
    return 0;

}

As you can see, my current solution is a little anemic. Thanks for the help, and I know everyone on the net doesn't necessarily like helping with other people's homework. ;P

Answer 1

char c[] = "This is an example of text for exercise (2).";
int d[26] = {0}, i, value;

for(i=0; i < sizeof(c) - 1; i++){ //-1 to exclude terminating NULL
   value = c[i]-'a';
   if(value < 26 && value >= 0) d[value]++;
}

for(i=0; i < 26; i++){
   if(d[i]) printf("Alphabet-%c Count-%d\n", 'a'+i, d[i]);
}

---

Corrected. Thanks caf and Leffler.

Answer 2

The intention of the question is for you to figure out how to efficiently convert a character between 'a' and 'z' into an index between 0 and 25. You are apparently allowed to assume ASCII for this (although the C standard does not guarantee any particular character set), which has the useful property that values of the characters 'a' through 'z' are sequential.

Once you've done that, you can increment the corresponding slot in your array d (note that you will need to initialise that array to all-zeroes to begin with, which can be done simply with char d[26] = { 0 };. At the end, you'd scan through the array d and print the counts out that are greater than zero, along with the corresponding character (which will involve the reverse transformation - from an index 0 through 25 into a character 'a' through 'z').

Answer 3

Fortunately for you, you do not seem to be required to produce a solution that would work on an EBCDIC machine (mainframe).

Your inner loop needs to be replaced by a conditional:

if (c[i] is lower-case alphabetic)
    increment the appropriate count in the d-array

After finishing the string, you then need a loop to scan through the d-array, printing out the letter corresponding to the entry and the count associated with it.

Your d-array uses 'char' for the counts; that is OK for the exercise but you would probably need to use a bigger integer type for a general purpose solution. You should also ensure that it is initialized to all zeros; it is difficult to get meaningful information out of random garbage (and the language does not guarantee that anything other than garbage will be on the stack where the d-array is stored).

Answer 4

.

for(s = 'a'; s < 'z'; s++){ 
    j=0;
    for(i = 0; i < sizeof(c); i++) { 
         if( c[i] == s )
            j++;
    }
    if (j > 0)
        printf("%c,%d\n", s, j); 
}

Answer 5

char c[] = "This is an example of text for exercise (2).";
char d[26];
int i;
int j;

for(i = 0; i < 26; i++)
{
    d[i] = 0; // Set the frequency of the letter to zero before we start counting.
    for(j = 0; j < strlen(c); j++)
    {
        if(c[j] == i + 'a')
            d[i]++;
    }
    if(d[i] > 0) // If the frequency of the letter is greater than 0, show it.
        printf("%c - %d\n", (i + 'a'), d[i]);
}