Question

I wanted to write a program which counts the occurrences of each letter in a string, then prints one of each letter followed by the count for that letter.

For example:

aabbcccd –
Has 2 a, 2 b, 3 c, and 1 d

So I’d like to convert and print this as:

a2b2c3d1

I wrote code (see below) to perform this count/conversion but for some reason I’m not seeing any output.

#include<stdio.h>
main()
{
    char array[]="aabbcccd";
    char type,*count,*cp=array;
    while(cp!='\0'){
      type=*cp;
      cp++;
      count=cp;
      int c;
      for(c=1;*cp==type;c++,cp++);
      *count='0'+c;
    }
    count++;   
    *count='\0';
    printf("%s",array);
}

Can anyone help me understand why I’m not seeing any output from printf()?

Answer 1

char array[]="aabbcccd";
char type,*count,*cp=array;
while(cp!='\0'){ 

*cp is a pointer it’s pointing to the address of the start of the array, it will never be == to a char '\0' so it can’t leave the loop.

You need to deference the pointer to get what it’s pointing at:

while(*cp != '\0') {
...

Also, you have a ; after your for loop, skipping the contents of it:

for(c=1;*cp==type;c++,cp++); <-- this ; makes it not execute the code beneath it

After fixing both of those problems the code produces an output:

mike@linux-4puc:~> ./a.out 
a1b1c2cd

Not the one you wanted yet, but that fixes your problems with “printf not functional”