How to automatically find how many times a character existed in a string?

Question

I'm a beginner in c++ and i made a code mentions how many times a character existed in a string :-

#include <string>
#include <iostream>
using namespace std;
int main(){
int s=0;
int l=0;
int o=0;
int w=0;
int y=0;
string boo="slowly";
for (size_t j = 0; j < boo.size(); j++) {
while (boo[j] == 's') {
s=s+1;
break;}
while (boo[j] == 'l')
{
l=l+1;
break;}
while (boo[j] == 'o') {
o=o+1;
break;}
while (boo[j] == 'w') {
w=w+1;
break;}
while (boo[j] == 'y') {
y=y+1;
break;}}

cout <<"s ="<<s<<endl;
cout <<"l ="<<l<<endl;
cout <<"o ="<<o<<endl;
cout <<"w ="<<w<<endl;
cout <<"y="<<y<<endl;
system("pause");
return 0;
}

I was wondering how to make a code that automatically detect the character in the string and apply the condition on it without making a while loop for every single alphabet character and making int variables for every character ?

**Excuse my bad Englihs

Possible duplicate of [Count character occurrences in a string](http://stackoverflow.com/questions/3867890/count-character-occurrences-in-a-string) — LogicStuff, Mar 05 '16 at 11:49
No sir i don't want exact character i want to automatically detect every character in the string and mention how many times every character repeated — Dreamer, Mar 05 '16 at 11:52

user764486 · Accepted Answer · 2016-03-05T12:08:06.950

1

You can use a map to keep track of the different character counts. You iterate through the input once, and for each character look it up in the map. If it doesn't already exist in the map, a new entry will be created with the value 0.

After you've created the map where each entry has the character as the key, and the number of occurrences as the value, you iterate through the map and print each entry.

#include <iostream>
#include <map>
#include <string>

int main()
{
    std::string input = "slowly";

    std::map<char, int> occurrences;

    for (char character : input)
    {
        occurrences[character] += 1;
    }

    for (auto& entry : occurrences)
    {
        std::cout << entry.first << '=' << entry.second << std::endl;
    }
}

edited Mar 05 '16 at 12:08

answered Mar 05 '16 at 12:02

user764486

160
5

It doesn't work it gives errors in this line for (auto character : input) – Dreamer Mar 05 '16 at 12:08
You need to make sure you are compiling for C++11 by using the flag ‘-std=c++11’ with your compiler. – user764486 Mar 05 '16 at 12:09

Sergey Kalinichenko · Answer 2 · 2016-03-05T12:09:50.507

Use a little trick. First, observe that all lower-case characters are encoded with numeric codes in ascending order:

      ASCII EBCDIC
'a' - 97    129
'b' - 98    130
'c' - 99    131
...
'z' - 122   169

The codes may not be consecutive, as in the case of EBCDIC, but that does not matter: make an int array of sufficient size, walk through your string, and for each lowercase character ch mark index ch-'a'. The expression represents the distance between the character code and the beginning of the alphabet.

Now you can walk the array of ints, and print the corresponding character by performing the inverse translation of index i:

 char ch = 'a'+i;

Here is a complete example:

string s;
getline(cin, s);
int seen['z'-'a'+1] {0};
for (int i = 0 ; i != s.size() ; i++) {
    char ch = s[i];
    if (!islower(ch))
        continue;
    seen[ch-'a']++;
}
for (int i = 0 ; i != 'z'-'a'+1 ; i++) {
    if (seen[i]) {
        cout << (char)('a'+i) << " - " << seen[i] << endl;
    }
}

Demo.

score 0 · Answer 3 · answered Mar 05 '16 at 12:00

Try this:

#include <string>
#include <iostream>
using namespace std;
int main(){
   int s=0,l=0,o=0,w=0,y=0;
   string boo="slowly";
   for (int j = 0; j < boo.size(); j++) {
       switch(boo.charAt(j)){
            case 's':
               s++;
               break;
            case 'l':
               l++;
               break;
            case 'o':
               o++;
               break;
            case 'w':
               w++;
               break;
            case 'y':
               y++;
               break;
       }
   }
   cout <<"s ="<<s<<endl;
   cout <<"l ="<<l<<endl;
   cout <<"o ="<<o<<endl;
   cout <<"w ="<<w<<endl;
   cout <<"y="<<y<<endl;
   system("pause");
   return 0;
}

I'm sorry, it is at(): switch(boo.at(j)){ : Let's change to this code. charAt is java function. C++ function is at. — jonghyon lee, Mar 05 '16 at 14:15

score 0 · Answer 4 · answered Mar 05 '16 at 12:06

If your compiler supports C++ 2011 then you can write the following way providing that letters will be listed in the order in which they are present in the string

#include <iostream>
#include <string>
#include <map>

int main()
{
    std::string s( "slowly" );
    auto comp = [&]( char c1, char c2 ) { return s.find( c1 ) < s.find( c2 ); };
    std::map<char, int, decltype( comp )> m( comp );

    for ( char c : s ) ++m[c];

    for ( auto p : m ) std::cout << p.first << ": " << p.second << std::endl;
}

The program output is

s: 1
l: 2
o: 1
w: 1
y: 1

Otherwise if the compiler does not support C++ 2011 then instead of the lambda you have to use a function class defined before main. Also youi should substitute the range-based for loop for an ordinary loop.

score 0 · Answer 5 · answered Mar 05 '16 at 12:48

0

I guess the algorithm-header of the STL will help you:

std::string boo="slowly";
std::cout << "s = " << std::count(boo.cbegin(), boo.cend(), 's') << std::endl;

answered Mar 05 '16 at 12:48

JVApen

11,008
5
31
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How to automatically find how many times a character existed in a string?

5 Answers5