Lambda capture as const reference?

Dib_241536 0
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Is it possible to capture by const reference in a lambda expression?

I want the assignment marked below to fail, for example:

#include <algorithm>
#include <string>

using namespace std;

int main()
{
    string strings[] = 
    {
        "hello",
        "world"
    };
    static const size_t num_strings = sizeof(strings)/sizeof(strings[0]);

    string best_string = "foo";

    for_each( &strings[0], &strings[num_strings], [&best_string](const string& s)
      {
        best_string = s; // this should fail
      }
    );
return 0;
}

Update: As this is an old question, it might be good to update it if there are facilities in C++14 to help with this. Do the extensions in C++14 allow us to capture a non-const object by const reference? (August 2015)

c++14 c++11 lambda c++
8 Answers

Using a const will simply have the algorithm ampersand set the string to it's original value, In other words, the lambda won't really define itself as parameter of the function, though the surrounding scope will have an extra variable... Without defining it though, it wouldn't define the string as the typical [&, &best_string](string const s) Therefore, its most likely better if we just leave it at that, trying to capture the reference.

user_93069270
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