Specified letter count with recursion - c

Question

I'm trying to write a recursive algorithm that obtains to count specified a letter by user. But, I'm stuck in two cases. First, I think I must get 2 as result, I can't. Second, If there is no limit key, for instance the limit character specified as z by user , how the characters can be scanned up to end character here g ? The problem little bit complicated for me. I need your advices and ideas. Thank you all appreciated answers.
example string is: how are you i am testing
another examples:
#include <stdio.h>
int lettercount(char* str, char key, char limit);
int main(){
char test[]="how are you i am testing";
int num;
num=lettercount(test,'a','t');
printf("%d",num);
return 0;
}
int lettercount(char* str, char key, char limit)
{
int count = 0;
if(str[0] == limit)
{
return 0;
}
else if(str[0] == key)
{
lettercount(&str[1], key, limit);
count++;
}
else
lettercount(&str[1], key, limit);
return count;
}

Answer 1

as the code is unwinding from the recursion(s)
it needs to accumulate the count
the following code should work for your needs.
Note: this returns 0 if key and limit are the same char
int lettercount(char* str, char key, char limit)
{
int count = 0;
if(str[0] == limit)
{
return 0;
}
// implied else, more char in string to check
if(str[0] == key)
{
count++;
}
count += lettercount(&str[1], key, limit);
return count;
} // end function: lettercount

Answer 2

With a recursive function, you need 3 things. (1) set up in the function that prepares for the next call; (2) a recursive call; and (3) a way to terminate the recursion. Here is one approach. Note: the version in the code below is a long version for readability, a short version is included at the end:
#include <stdio.h>
/* recursively find the number of occurrences
of 'c' in 's' (n is provided as '0')
*/
int countchar (char *s, char c, int n)
{
char *p = s;
if (!*p)
return n;
if (*p == c)
n = countchar (p+1, c, n+1);
else
n = countchar (p+1, c, n);
return n;
}
int main (int argc, char **argv) {
if (argc < 3) {
fprintf (stderr, "\n error: insufficient input. Usage: %s <string> <char>\n\n", argv[0]);
return 1;
}
int count = countchar (argv[1], *argv[2], 0);
printf ("\n There are '%d' '%c's in: %s\n\n", count, *argv[2], argv[1]);
return 0;
}
output:
$ ./bin/rec_c_in_s "strings of s'es for summing" s
There are '5' 's's in: strings of s'es for summing
You can make the function shorter, but slightly less readable with:
int countchar (char *s, char c, int n)
{
char *p = s;
if (!*p) return n;
return countchar (p+1, c, (*p == c) ? n+1 : n);
}