Need help: Using Recursion function to count a letter

[jinx]

I finally figured it out

Code: Select all

#include<iostream.h>
#include<string.h>

int word(char*);

void main()
{


   char w[30];


   cout << "Please enter a word: ";
   cin  >> w;



   cout << endl <<"We have " << word(w) << " A's on this word" << endl << endl;





}


int word(char *x)
{
   int ctr = 0;

   if( *x != '\0')
   {
      if( *x != 'a')
      {
         x++;
         return word(x);
      }
      else
      {
   
         ctr++;
      
         x++;
         return ctr + word(x);
      
      }

   }
   else
   {
      return 0;
   }



}

[Alvaro]

Code: Select all

int w(char *x){return *x?w(x+1)+(*x=='a'):0;}

[jinx]

alvaro your code is very simple....i tried it and it works the same as my code. I have a question about the "?" on your code. What is that for? , i guess i havent learn that yet.

[WaltP]

It's a shorthand if/else

Code: Select all

if (a == 3)
{
    b = 5;
}
else
{
    b = 10;
}


is written as

Code: Select all

b = (a==3)? 5 : 10;


IOW
comparison ? value if true : value if false;

[Alvaro]

The other tricky part in my function is that using as an integer, (*x=='a') will evaluate to 1 when it's true and to 0 when it's false.

[WaltP]

The other tricky part in my function is that using as an integer, (*x=='a') will evaluate to 1 when it's true and to 0 when it's false.

Tricky? Why give a noob a tricky answer that he can't understand, instead of a straight forward one he can?

Showoff

[Alvaro]

Why give a noob a tricky answer that he can't understand, instead of a straight forward one he can?

Ok, the non-tricky version of that function is this:

Code: Select all

unsigned count_a(const char *s){
   if(*s=='\0')
      return 0;
   if(*s=='a')
      return 1+count_a(s+1);
   else
      return count_a(s+1);
}

Showoff

Just a little bit.

[jinx]

very nice....thanks guys