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Access captured variables outside the lambda


Access captured variables outside the lambda

By : Acoidán besay Glez R
Date : October 25 2020, 07:10 AM
I hope this helps . This is not possible. The lambda is a black box.
Option 1: use the return value
code :
template <typename F>
void bar(const F& f)
{
    auto d = f();
}
int main()
{
    int x;
    bar([&]{
        auto y=x;  // in case do_something would change x
        do_something(x);
        return y; 
    });
}
template <typename F>
void bar(const F& f)
{
    auto d = f();
    cout << d.first<<endl;  
}
int main()
{
    int x;
    bar([&]{
        auto y=x; 
        do_something(x);
        return make_pair(y, 0); // first is parameter, second real return
    });
}
template <typename F>
void bar(const F& f)
{
    auto d = f.x;
    f();
}
int main()
{
    struct Tmp {  // I won't need this callable elsewhere... 
        int &x; 
        Tmp (int& x) : x(x) {}
        void operator() () const {
           do_something(x);
        }
    };

    int x;
    Tmp t(x); 
    bar(t);
}


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C++ Lambda: Access static method in lambda leads to error 'this was not captured for this lambda function'

C++ Lambda: Access static method in lambda leads to error 'this was not captured for this lambda function'


By : Martin Kopecky
Date : March 29 2020, 07:55 AM
help you fix your problem Consider the following code: , The following work-around works:
code :
template< typename T > T* global_instance() { return T::instance(); }

void(*func_pointer)(void) = []{
    global_instance<Singleton>()->bar();
};
Get captured variables from lambda?

Get captured variables from lambda?


By : Sergio Ricardo Gomes
Date : March 29 2020, 07:55 AM
it fixes the issue I was wondering, if there's a way to get the types/values of the captured variables of a lambda? - The usage scenario would be something alike; , It's not possible by design
Is it possible to access (read only) the variables captured by a lambda?

Is it possible to access (read only) the variables captured by a lambda?


By : Marian Leonard
Date : March 29 2020, 07:55 AM
it fixes the issue Is it possible to access (read only) the variables captured by a lambda?
code :
auto plus( double a ) {
  using R = struct {
    double a;
    double operator()(double b)const{return b+a;}
  };
  return R{std::move(a)};
}
C++: Change variables captured in lambda from outside

C++: Change variables captured in lambda from outside


By : Elena Maksimovich
Date : March 29 2020, 07:55 AM
wish help you to fix your issue To answer your question, no it is not possible to change local variables of a function/functor (which is what lambda is) from the outside.
Think of lambda as "regular" functions whose name you don't have. Just like for regular functions you can't change local variable values from the outside (params passed by value), you can't do it with lambda too. This is a necessary requirement for security & predictability of the result.
Difference between ref-captured and non-explicitly captured constexpr variables in lambda-expressions

Difference between ref-captured and non-explicitly captured constexpr variables in lambda-expressions


By : dedard9fx
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , If we use clang as a compiler, which is usually more relevant than gcc when it comes to language lawyer, we find out that a simplified example is very telling:
code :
template<typename T>
int foo(T/*&*/ l) {
    constexpr auto x = l() - 1;

    if constexpr(x <= 0) {
        return 42;
    }
    else {
        return 0;
    }
}

auto l2 = []{
    return 3;
};

int main() {
    foo(l2);
}
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