Counting Alphabetic Characters That Are Contained in an Array with C
- by Craig
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
