const std::unordered_map’ as ‘this’ argument discards qualifiers in lambda

Question

I am now sorting a string in decreasing order based on the frequency of characters. For example: the input is "apple", I must output "ppale". Because character 'p' appears twice and the others only appear once. so 'p' must be placed to the left most.

Here is my code:

string frequencySort(string s) {
    if(s.size() == 0)   return "";

    unordered_map<char,int> map;
    for(auto c : s)
        map[c]++;   //record the frequency of each characters

    sort(s.begin(),s.end(),
        [=](char a, char b){return map[a] > map[b] || map[a]==map[b]&&a<b;}
        );
    return s;
}

But compiler shows error: passing ‘const std::unordered_map<char, int>’ as ‘this’ argument discards qualifiers on [=](char a, char b){return map[a] > map[b] || map[a]==map[b]&&a<b;}.

However, if I capture local variable by reference, i.e.

[&](char a, char b){return map[a] > map[b] || map[a]==map[b]&&a<b;}

it works.

Could anyone please tell my why I cannot capture by value?

Answer 1

A lambda's operator (...) is const by default. Capture by value will require the lambda to keep copies of all captured objects.

Make the lambda mutable to allow modifying value captures.

sort(s.begin(),s.end(), [=](char a, char b) mutable 
          {return map[a] > map[b] || map[a]==map[b]&&a<b;}
    );

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I suspect the sort and unordered_map is from the namespace std. From the way you used them without a namespace resolution, you probably have a using namespace std in your code (which isn't nice). I will be weary of using map for a variable name.