Recursive function for "vowels",Recursive function for "sum of array"

[srikantk07]

Hello,

Will you please write a c++ program which includes all functions,

1) Recursive function for "vowels", that returns number of vowels in string.

2) Recursive function for "sum of array"

3) Program that uses a recursive function to check whether a string is Palindrome

4) Program that uses recursive function to print string backward

5) write a Recursive function for "backword number"(i.e: reversedigits, that takes an integer as a parameter and returns the number with digits)

6) Recursive function for "power"

Please let me know if i am not clear.

Thank you so much

[Smac89]

#include <iostream>
#include <cmath>
#include <algorithm>

using std::cout;
using std::cin;
using std::endl;
using std::string;

struct Words
{
  int CountVowels(string, int);
public:
  Words(string);
  bool isPal(string, int);
  void Print();
  string Skippy;
  int numVowel;
  int length;
  ~Words();
};


Words::Words (string m)
{
  Skippy = m;
  length = 0;
  numVowel = CountVowels(Skippy, 0);
}

int Words::CountVowels(string myWord, int startindex)
{
  length++;
  int pandi;
  
  std::transform(myWord.begin(), myWord.end(), myWord.begin(), tolower);
  
  if (myWord[startindex])
  {
    if (myWord[startindex] != 'a' && myWord[startindex] != 'e' && myWord[startindex] != 'i' && myWord[startindex] != 'o' 
      && myWord[startindex] != 'u')
      pandi = 0;
    else
      pandi = 1;
    return pandi + CountVowels(myWord, startindex+1);
  }    
    return 0;
}



bool Words::isPal(string myWord, int size)
{
  int r = myWord.size() - size;
  int t = size - 1;
  
  std::transform(myWord.begin(), myWord.end(), myWord.begin(), tolower);
  
  if (size == r || r == t)
    return true;
  
  //size = r is true when it is an even size string and 
    //the 2 middle characters have been checked
    
    //r = t is true when it is an odd size string and the 
    //two chars on either side of the middle char have been checked
    
    if ((myWord[r]) != (myWord[t]))
      return false;
    
    return isPal(myWord, --size);
}

void Words::Print()
{
  cout << Skippy[--length];
  if (length == 0)
  {
    cout <<endl;
    return;
  }
  Print();
}

Words::~Words()
{
  cout << "\ntilde delete\n\n";
}


long pows_n(long val, int exp)
{
  if (exp == 0)
    return 1;
  return val * pows_n(val, --exp);
}


void BackwardNum(long num, bool init = true)
{
  if (log10(num) > 0)
  {
    int t = num/10;
    int k = t * 10;
    cout << (num - k);
    num = t;
    BackwardNum(num, init = false);
  }
  else
  {
    if (num > 0 || init)
      cout << num;
    cout << endl;
  }
}


int main()
{
  string test;
  long Val;
  cout << "\nEnter a string: ";
  getline(cin, test);
  
    Words myTestWord(test);
    cout << "The string " << myTestWord.Skippy << " is " << 
    (myTestWord.isPal(myTestWord.Skippy, myTestWord.Skippy.size()) ? "palindrome\n" : "not palindrome\n");
    
    cout << "The number of vowels in the string is " << myTestWord.numVowel << endl;
    
    cout <<"String backwards is ";
    myTestWord.Print();    
    
    cout << "\nEnter a number: ";
    cin >> Val;
    
    cout << Val << " to the power of 2 is " << pows_n(Val, 2);
    cout << "\nNumber backwards is ";
    BackwardNum(Val);
    
  return 0;
}

Enter a string: What is yours sruoy si tAHW
The string What is yours sruoy si tAHW is palindrome
The number of vowels in the string is 8
String backwards is WHAt is yours sruoy si tahW

Enter a number: 2563
2563 to the power of 2 is 6568969
Number backwards is 3652

tilde delete

[vlad from moscow]

@Smac89

Your code is invalid. For example function BackwardNum does not return a reverse number. And I think that it shall be declared as

long BackwardNum( long );

instead of

void BackwardNum(long num, bool init = true)

that is without any additional parameter except the parameter that will get the original number.

[Smac89]

I see, here is one that works well

long BackwardNum(long num)
{
  int t = log10(num);
  if (t == 0)
    return num;
  return (((num%10) * pows_n(10, t)) + BackwardNum(num/10));
}