Count decimal numbers in a string using the standard library

user_8974764 0
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I have white-space separeted integers inside a string, e.g.:

std::string s = "1 2 33 444 0 5";

The string is well-formed: just white-space separated numbers without any letters, new-lines, etc.

How to count the number of integers in the above string in an STL way? I am looking for something "short" that will use e.g. iterators or <algorithm>.

c++
5 Answers

std::count_if(s.begin(),s.end(), [](unsigned char c){ return std::isspace(c);}) + 1

Edit:

If there are multiple/varying spaces between chars, then you can change the lambda to:

    [](unsigned int c)
    { 
        static bool prev = false;
        bool current = isspace(c);
        bool new_space = !prev && current;
        prev = current;
        
        return new_space;
    }
user_34769540

A simple way is to use string streams:

#include <iostream>
#include <sstream>

int main()
{
    int temp;
    int count = 0;
    std::string s = "1 2 33 444 0 5";

    std::stringstream ss(s);
    while(ss >> temp){
        count++;
    }
    std::cout << count; //test print
    return EXIT_SUCCESS;
}

Note that this will only count parseabe values, for instance, if a non-numeric character is found (appart from spaces) it will stop counting. It also works for multiple spaces.

user_68659320

This solution uses only the STL, has no loops, and will handle arbitrary number of leading, trailing, and extra whitespace:

std::string s = "1 2 33 444 0 5";
std::stringstream ss(s);

int const count = std::distance(std::istream_iterator<int>{ss},
                                std::istream_iterator<int>{});

Here's a demo.

user_83728530

Are the integers guaranteed to each be separated by a single whitespace character (without there being leading or trailing whitespace)?

If so, then you just need to add 1 to the number of whitespace chars in the string itself, accounting for the empty string being a special case.

#include <algorithm>

size_t intCount(const std::string& s) {
  if (s.size() == 0) {
    return 0;
  }

  return std::count(s.begin(), s.end(), ‘ ‘) + 1;
}
user_136534180

You can use regular expressions:

#include <iterator>
#include <regex>
#include <string>

std::string s = "1 2 33 444 0 5";

const std::regex regex("\\d+");
const auto n = std::distance(
    std::sregex_iterator(s.begin(), s.end(), regex),
    std::sregex_iterator());

This will handle multiple, leading and trailing whitespaces automatically. This approach doesn't impose any limitation on the length of each integer.

user_16251870
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