使用 string.substr(post,len) 的无限循环

Question

Could you figure out why it keeps looping infinitely in the console? The programmer's supposed to list out each character of a user-inserted string and next to each unique character, in brackets, it's supposed to display the number of times that character occurs in the string... no idea why.

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

int main () {
  string input;
  cout << "input string: " , cin >> input;
  sort (input.begin() , input.end());
  while (!input.empty()) {
    int j{1}, i{0};
    while (input.at(i) == input.at(i+1)) {
      j++;
      i++;
    }
    cout << input.at(i) << " (" << j << "), ";
    input.substr(i);
  }
  return 0;
}

Answer 1

This statement

input.substr(i);

does not change the object input itself.

So, either you will have an infinite loop if for some index i input.at(i) is not equal to input.at(i+1) , or you can have an exception out of range because i + 1 can be equal to input.size() .

From the description of the member function at

Throws: out_of_range if pos >= size().

The program can be implemented in different ways. For example the following way

#include <iostream>
#include <string>
#include <algorithm>

int main() 
{
    std::cout << "input string: ";
    std::string input;
      
    std::cin >> input;
      
    std::sort( input.begin() , input.end() );
    
    for ( size_t i = 0; i < input.size(); )
    {
        size_t j = input.find_first_not_of( input[i], i );
        
        if ( j == std::string::npos ) j = i + 1;
        
        if ( i != 0 ) std::cout << ", ";
        std::cout << input[i] << " (" << j - i << ")";

        i = j;
    }
    
    std::cout << '\n';
    
    return 0;
}

The program output is

input string: Hello
H (1), e (1), l (2), o (1) 

Or you can use the standard container std::map or std::unordered_map as for example

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

int main() 
{
    std::cout << "input string: ";
    std::string input;
      
    std::cin >> input;
      
    std::map<char, size_t> m;
    
    for ( const auto &c : input )
    {
        ++m[c];
    }

    bool first = true;
    for ( const auto &p : m )
    {
        if ( !first  ) std::cout << ", ";
        std::cout << p.first << " (" << p.second << ")";
        first = false;
    }
    
    std::cout << '\n';
    
    return 0;
}

If you want that characters of the inputted string were output in the order in which they are present in the string then the program can look like

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

int main() 
{
    std::cout << "input string: ";
    std::string input;
      
    std::cin >> input;
      
    auto less = [&input]( const auto &c1, const auto &c2 )
    {
        return input.find( c1 ) < input.find( c2 );
    };
    
    std::map<char, size_t, decltype( less )> m( less );
    
    for ( const auto &c : input )
    {
        ++m[c];
    }
    
    bool first = true;
    
    for ( const auto &p : m  )
    {
        if ( !first  ) std::cout << ", ";
        std::cout << p.first << " (" << p.second << ")";
        first = false;
    }
    
    std::cout << '\n';
    
    return 0;
}

Or without changing the original string and without using an additional container the program can look the following way.

#include <iostream>
#include <string>

int main() 
{
    std::cout << "input string: ";
    std::string input;
      
    std::cin >> input;
      
    for ( size_t i = 0; i < input.size(); i++ )
    {
        size_t j = 0;
        
        while ( j != i && input[j] != input[i] ) j++;
        
        if ( j == i )
        {
            size_t count = 1;
            while ( ++j < input.size() )
            {
                if ( input[j] == input[i]  ) ++count;
            }
            if ( i != 0  ) std::cout << ", ";
            std::cout << input[i] << " (" << count << ")";
        }
    }
    
    std::cout << '\n';
    
    return 0;
}

The program output might look like

input string: elephant
e (2), l (1), p (1), h (1), a (1), n (1), t (1)