C ++ 'da bir dizede karakter oluşumlarını sayma

"_"Bir dizedeki sayısını nasıl sayabilirim "bla_bla_blabla_bla"?

#include <algorithm>

std::string s = "a_b_c";
size_t n = std::count(s.begin(), s.end(), '_');

pseudocode:

count = 0
For each character c in string s
  Check if c equals '_'
    If yes, increase count

EDIT: C++ example code:

int count_underscores(string s) {
  int count = 0;

  for (int i = 0; i < s.size(); i++)
    if (s[i] == '_') count++;

  return count;
}

Bu kodun, std::string, if you're using char*, replace s.size() with strlen(s).

Also note: I can understand you want something "as small as possible", but I'd suggest you to use this solution instead. As you see you can use a function to encapsulate the code for you so you won't have to write out the for loop everytime, but can just use count_underscores("my_string_") in the rest of your code. Using advanced C++ algorithms is certainly possible here, but I think it's overkill.

Uygun şekilde adlandırılmış değişkenlerle eski moda çözüm. Bu koda biraz ruh verir.

#include <cstdio>
int _(char*__){int ___=0;while(*__)___='_'==*__++?___+1:___;return ___;}int main(){char*__="_la_blba_bla__bla___";printf("The string \"%s\" contains %d _ characters\n",__,_(__));}

Düzenleme: yaklaşık 8 yıl sonra, bu cevaba bakarak ben bunu yaptım utanıyorum (düşük çaba sorusunda sinsi bir dürtmek olarak kendime haklı olsa bile). Bu toksiktir ve TAMAM değildir. Gönderiyi kaldırmıyorum; StackOverflow atmosfer değiştirmeye yardımcı olmak için bu özür ekliyorum. O halde OP: Özür dilerim ve umarım trollingime rağmen ödevini doğru bulmuşsundur ve benimki gibi cevaplar siteye katılmanı cesaretlendirmedi.

#include <boost/range/algorithm/count.hpp>

std::string str = "a_b_c";
int cnt = boost::count(str, '_');

Siz adlandırın ... Lambda versiyonu ... :)

using namespace boost::lambda;

std::string s = "a_b_c";
std::cout << std::count_if (s.begin(), s.end(), _1 == '_') << std::endl;

Birkaç içeriğe ihtiyacınız var ... Bunu bir egzersiz olarak bırakıyorum ...

Using the lambda function to check the character is "_" then only count will be incremented else not a valid character

std::string s = "a_b_c";
size_t count = std::count_if( s.begin(), s.end(), []( char c ){if(c =='_') return true; });
std::cout << "The count of numbers: " << count << std::endl;

There are several methods of std::string for searching, but find is probably what you're looking for. If you mean a C-style string, then the equivalent is strchr. However, in either case, you can also use a for loop and check each character—the loop is essentially what these two wrap up.

Once you know how to find the next character given a starting position, you continually advance your search (i.e. use a loop), counting as you go.

Count character occurrences in a string is easy:

#include <bits/stdc++.h>
using namespace std;
int main()
{
    string s="Sakib Hossain";
    int cou=count(s.begin(),s.end(),'a');
    cout<<cou;
}

You can find out occurrence of '_' in source string by using string functions. find() function takes 2 arguments , first - string whose occurrences we want to find out and second argument takes starting position.While loop is use to find out occurrence till the end of source string.

example:

string str2 = "_";
string strData = "bla_bla_blabla_bla_";

size_t pos = 0,pos2;

while ((pos = strData.find(str2, pos)) < strData.length()) 
{
    printf("\n%d", pos);
    pos += str2.length();
} 

Bu şekilde yapardım:

#include <iostream>
#include <string>
using namespace std;
int main()
{

int count = 0;
string s("Hello_world");

for (int i = 0; i < s.size(); i++) 
    {
       if (s.at(i) == '_')    
           count++;
    }
cout << endl << count;
cin.ignore();
return 0;
}

Böyle bir şey yapardım :)

const char* str = "bla_bla_blabla_bla";
char* p = str;    
unsigned int count = 0;
while (*p != '\0')
    if (*p++ == '_')
        count++;

Deneyin

#include <iostream>
 #include <string>
 using namespace std;


int WordOccurrenceCount( std::string const & str, std::string const & word )
{
       int count(0);
       std::string::size_type word_pos( 0 );
       while ( word_pos!=std::string::npos )
       {
               word_pos = str.find(word, word_pos );
               if ( word_pos != std::string::npos )
               {
                       ++count;

         // start next search after this word 
                       word_pos += word.length();
               }
       }

       return count;
}


int main()
{

   string sting1="theeee peeeearl is in theeee riveeeer";
   string word1="e";
   cout<<word1<<" occurs "<<WordOccurrenceCount(sting1,word1)<<" times in ["<<sting1 <<"] \n\n";

   return 0;
}

public static void main(String[] args) {
        char[] array = "aabsbdcbdgratsbdbcfdgs".toCharArray();
        char[][] countArr = new char[array.length][2];
        int lastIndex = 0;
        for (char c : array) {
            int foundIndex = -1;
            for (int i = 0; i < lastIndex; i++) {
                if (countArr[i][0] == c) {
                    foundIndex = i;
                    break;
                }
            }
            if (foundIndex >= 0) {
                int a = countArr[foundIndex][1];
                countArr[foundIndex][1] = (char) ++a;
            } else {
                countArr[lastIndex][0] = c;
                countArr[lastIndex][1] = '1';
                lastIndex++;
            }
        }
        for (int i = 0; i < lastIndex; i++) {
            System.out.println(countArr[i][0] + " " + countArr[i][1]);
        }
    }