repeat in a string

[rezy3312]

So I am writing a code where i want to print out the amount of letters in a string and print it out once but have minor problems of reading the previous letter. For example if the user enters " hello there class" it should print out to the user

There are 2 "h"
there are 3 "e"
there are 3 "l"....etc

can someone help me out and show me the way i am pretty sure I use a pointer but is it possible not to use a pointer because i get really confused when using them. Thank you in advanced.

  


#include <iostream>
#include <cctype>

using namespace std;

int main()
{
    string sentence = " ";
    
    cout << "enter sentence to see how many letter of each you will get " << endl;
    getline(cin,sentence);
    
    cout << "here is the senctence you entered\n\n"  << sentence << endl;
    
    for(int i = 0; i < sentence.length(); i++){
        
        if(sentence.at(i) != ' ' || sentence.at(i) != '\n'){
            if(isalpha(sentence[i])){
                cout << "there are " << i << "letter of " << sentence.at(i) << endl;
            }
            else if(isdigit(sentence[i])){
                cout << "there are " << i << "numbers listed" <<sentence.at(i) << endl;
            }
            else{
                if(isspace(sentence[i])){
                    continue;
                }
            }
        }
        
        else{
            cout << "error occured" << endl;
        }
    }

    return 0;
}

[jonnin]

favor [] over .at(); its faster, only use .at() if you may go out of bounds (it checks).

there are 2 fairly straightforward ways to do this.
1) sort a copy of the string, and iterate, it will be in a form abbbcddde etc where you can just count the same ones until it changes to something else. the length will be the same as it was.

2) you can make a container (vector, or even a c-array) of integers set to zero and increment each one by using the letter's numeric value (this is not useful in unicode, but its fine for ascii):
char counter[256] = {0};
for(...)
counter[string[i]]++;

cout << blah bah there are << counter['e'] << e's in that.

you don't need any pointers at all for either approach.
don't forget that 'e' and 'E' are not the same, if you care.

[rezy3312]

yah that makes sense in a way i kinda understand so if i create a empty string and add the element once to that string i can compare it if i understand it right sort of like this in a way but of course there are some syntex errors

//for loop begin

copy[i] = sentence[i];

if(copy[i] != copy[i + 1]){
                cout << "the letter is " << copy[i]  << "has already been read" << endl;
                
}
else{
      if(copy[i] == sentence[i]){
            cout << "The letter " << copy[i] << "has not been read" << endl;
                 continue;
      }
}

//for loop end

[jonnin]

I have no idea what you are doing there.
if(copy[i] == sentence[i]){ //this is always true, you never modified them and its a copy of it.

if you are trying approach 1) you forgot the SORT part?
assuming a string has been sorted, then, its more like

tmp = sorted[0];
count = 1;
for(all the letters)
{
if(sorted[i] == sorted[i+1]_
count ++;
else
{
cout "words" there are count number of sorted[i]'s found
count = 1; //after printing it! not before :P
//here it resets the counter and next iteration is checking the i+1 that was different against i+2 .. new letter, new count... etc..
}
}

[againtry]

@OP
Here's approach no. 2 using as much as your code as possible ...

#include <iostream>
#include <cctype>

using namespace std;

int main()
{
    int character_count[26]{0};
    
    string sentence = " ";
    
    cout << "enter sentence to see how many letter of each you will get: ";
    getline(cin,sentence);
    
    cout << "here is the senctence you entered\n"  << sentence << endl;
    
    char ch;
    for(int i = 0; i < sentence.length(); i++)
    {
        ch = toupper(sentence[i]);
        
        if( isalpha(ch) )
        {
            character_count[ch - 'A']++;
        }
    }
    
    for (int i = 0; i < 26; i++)
    {
        cout
        << "there are " << character_count[i] << " letter of "
        << (char)('A' + i) <<  endl;
    }
    
    return 0;
}

[jonnin]

you may want to add
if(character_count[i])
at line 30 so you don't print all the zero count guys. Your call.

[againtry]

@jonnin Good call - I was lazy.
@OP please get on with making the necessary changes. Then say thanks to us and green tick so we can all move on :)