Is this lambda capturing problem a gcc compiler bug?

Minimal working example:

#include <iostream>
#include <memory>
#include <string>

int main()
{
    std::shared_ptr<std::string> i = std::make_shared<std::string>("foo");

    auto f = [=]()
        {
            i.reset();
            std::cout << i.get() << "\n";
        };

    std::cout << i.use_count() << "\n";
    f();
    std::cout << i.use_count() << "\n";
}

      

Compiler error:

$ g++ -std=c++11 /tmp/foo.cpp 
/tmp/foo.cpp: In lambda function:
/tmp/foo.cpp:11:12: error: passing ‘const std::shared_ptr<std::basic_string<char> >’ as ‘this’ argument of ‘void std::__shared_ptr<_Tp, _Lp>::reset() [with _Tp = std::basic_string<char>; __gnu_cxx::_Lock_policy _Lp = (__gnu_cxx::_Lock_policy)2u]’ discards qualifiers [-fpermissive]
    i.reset();

      

I believe it i

should be captured as a value, but it seems to be perceived as a const value.

Compiler version:

g++ (GCC) 4.9.2 20141101 (Red Hat 4.9.2-1)

      

+3


source to share


1 answer


shared_ptr

is a member of the closure object. And operator()

marked const

.
Thus, you cannot change i

, i.e. Call functions const

not const

like reset

.

Try



auto f = [=]() mutable
{
    i.reset();
    std::cout << i.get() << "\n";
};

      

+7


source







All Articles