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When two C-strings have most frequent letter.

dasboshiet (30)
I am confused why when I enter for example... sese, I get E as the string (I understand because that appears first) but I get A as the c-string.

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#include <iostream>
#include <cstring>
#include <string>
#include <vector>

using namespace std;

//Function prototypes
char mostfreq(string);
char mostfreq(char []);

int main()
{
    string words;
    char freq;
    //Generates 100 char space
    char letters[100];

    /*************************************************************************/
    /*******************************STRING************************************/
    cout << "Enter anything you want and I will tell you the most frequent letter." << endl;
    getline(cin, words);

    freq = mostfreq(words);
    cout << "The most frequent letter is " << freq << " in a string." << endl;

    /*************************************************************************/
    /******************************C-STRING***********************************/

    for(int i = 0; i < words.length(); i++)
    {
        letters[i] = words[i];
    }

    freq = mostfreq(letters);
    cout << "The most frequent letter is " << freq << " in a c-string." << endl;
}

char mostfreq(string words)
{
    //Tally mark for each letter that appears
    int wordTally[26];
    int highest = 0;

    //Sets Tally Marks to 0
    for(int i = 0; i < 26; i++)
    {
        wordTally[i] = 0;
    }

    //Changes all words to upper case
    for(int j = 0; j < words.length(); j++)
    {
        if(islower(words[j]))
        {
            words[j] = toupper(words[j]);
        }
    }

    //Subtracts 65(ASCII Decimal for 'A') and uses that number to mark on Tally
    for(int k = 0; k < words.length(); k++)
    {
        words[k] = words[k] - 65;
        wordTally[words[k]]++;
    }

    //Determines which Tally haves the highest number and will be recorded
    for(int l = 1; l < 26; l++)
    {
        if(wordTally[highest] < wordTally[l])
        {
            highest = l;
        }
    }

    //The Tally with the highest count will be returned
    //Add 65 because it was subtracted earlier
    return highest + 65;
}

char mostfreq(char letters[])
{
    int wordTally[26];
    int highest = 0;
    int counter = 0;
    //Vector is used to hold all the letters from the given word
    std::vector <char> wordCheck;

    for(int i = 0; i < 26; i++)
    {
        wordTally[i] = 0;
    }

    //Use vector to avoid all non alphabet characters
    while(letters[counter] != '\0')
    {
        if(isalpha(letters[counter]))
        {
            wordCheck.push_back(letters[counter]);
        }
        counter++;
    }

    //Change all lower cases to upper case
    for(int j = 0; j < wordCheck.size(); j++)
    {
        if(islower(wordCheck[j]))
        {
            wordCheck[j] = toupper(wordCheck[j]);
        }
    }

    //Subtracts 65 to determine where to mark on tally
    for(int k = 0; k < wordCheck.size(); k++)
    {
        wordCheck[k] = wordCheck[k] - 65;
        wordTally[wordCheck[k]]++;
    }

    //Searches for the highest mark on Tally
    for(int l = 1; l < wordCheck.size(); l++)
    {
        if(wordTally[highest] < wordTally[l])
        {
            highest = l;
        }
    }

    //Returns the most frequent letter
    return highest + 65;
}



Here is the result...
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Enter anything you want and I will tell you the most frequent letter.
sese
The most frequent letter is E in a string.
The most frequent letter is A in a c-string.

Last edited on
AbstractionAnon (6954)
Lines 30-33: You never store a \0 terminator into the C-string. mostfreq(char[]) is indexing past the end of the entered string.
Peter87 (11179)
You don't loop through all 26 letters in the last loop.
Last edited on
dasboshiet (30)
how do i store the \0?
dasboshiet (30)
ah... I just realized I made an error on that. Thanks for pointing that out Peter! it works now
dasboshiet (30)
does 
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    for(int i = 0; i < words.length(); i++)
    {
        if(i != words.length())
            letters[i] = words[i];
        else if(i == words.length())
            letters[i+1] = '\0';
    }

include the \0 ?
AbstractionAnon (6954)
does <snip> include the \0 ?

No. words.length() does not include the trailing \0.

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    for(int i = 0; i < words.length(); i++)
        letters[i] = words[i];
    letters[words.length()] = '\0';





dasboshiet (30)
makes sense, thanks a lot!
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