count vowels in a string with recursion function - c++

Question

I am currently in my last year of a BS in Information Systems with a major in programming. I took and passed C++ programming 1. I am now in C++ programming 2 and am having trouble with recursive functions. We have a homework assignment, where we are suppose to write a program that will count the number of vowels in a string that is entered by the user.
I have a program similar to this from my C++ programming 1 class that works using a for loop and if-then statements. I had assumed that it would be easy to convert this working program to use a recursive function, I was wrong.
I have the code (not looking for someone to do it for me) and I think i have it setup right. Just am not sure where to put the call to the function within the function.
Can somebody point me in the right direction?`
This is my first time asking a question. If i attached my code wrong, please let me know.
#include <iostream>
#include <string>
#include <cctype>
using namespace std;
int vowelCount(string, int, int&);
int mail()
{
string input;
int len;
//int x;
//int y;
int count;
count = 0;
cout << "Enter a string of characters with no spaces: ";
cin >> input;
len = input.length();
vowelCount(input, len, count);
cout << "There were " << count << " vowels." << endl;
system("pause");
return 0;
}
int vowelCount(string input, int len, int& count)
{
int y;
int x;
y = input.at(len);
if (len == 1)
{
count = count + 1;
return count;
}
else
{
y = input.at(len);
if ((y == 'a') || (y == 'e') || (y == 'i') || (y == 'o') || (y == 'u') || (y == 'A') || (y == 'E') || (y == 'I') || (y == 'O') || (y == 'U'))
{
count = count + 1;
len = len - 1;
}
else
{
len = len - 1;
vowelCount(string input, int len, int& count);
return count;
}
}
}
return 0;
}

Answer 1

For a general understanding, I recommend the answers to this question.
First of all, this code doesn't run: it has syntax errors. You shouldn't be asking for logic help until you have a program that at least runs. You cannot call a function with the entire signature. FOr instance, the final block should be a simple
return vowelCount(input, len-1)
You're returning the count as both the function's value and a parameter. Drop the parameter.
Now, for understanding the recursion ... take this in a couple of steps:
If the string is empty, return 0.
Otherwise, check the current letter:
2T if it's a vowel, return 1 + {count on rest of string}
2F else, return {count on rest of string}
Your two recursive calls are in the braces. Can you take it from here?

Answer 2

I would write the function the following way
#include <iostream>
#include <string>
#include <cstring>
#include <cctype>
std::string::size_type vowelCount( const std::string &input, std::string::size_type pos = 0 )
{
const char *vowels = "AEIOU";
return pos >= input.size()
? 0
: ( std::strchr( vowels, std::toupper( ( unsigned char )input[pos] ) ) != nullptr )
+ vowelCount( input, pos + 1 );
}
int main()
{
std::string s;
std::cin >> s;
std::cout << "There were " << vowelCount( s ) << " vowels." << std::endl;
return 0;
}
If for example to enter
AaBbCcDdEe
then the output will be
There were 4 vowels.
I suppose that the string does not contain embedded zero characters.:)
Otherwise you should substitute the condition
( std::strchr( vowels, std::toupper( ( unsigned char )input[pos] ) ) != nullptr )
for
( input[pos] != '\0' && std::strchr( vowels, std::toupper( ( unsigned char )input[pos] ) ) != nullptr )
As for your function then if to write it syntactically valid it does not make sense due to for example this statement
int vowelCount(string input, int len, int& count)
{
int y;
int x;
y = input.at(len);
^^^^^^^^^^^^^^^^^^
because according to the C++ Standard member function at
5 Throws: out_of_range if pos >= size().