Otdfbt

In the below program, when mutable is not used, the program fails to compile.

#include <iostream>
#include <queue>
#include <functional>

std::queue<std::function<void()>> q;

template<typename T, typename... Args>
void enqueue(T&& func, Args&&... args)
{
 //q.emplace([=]() // this fails
 q.emplace([=]() mutable //this works
 func(std::forward<Args>(args)...);
 );


int main()

 auto f1 = [](int a, int b) std::cout << a << b << "n"; ;
 auto f2 = [](double a, double b) std::cout << a << b << "n";;
 enqueue(f1, 10, 20);
 enqueue(f2, 3.14, 2.14);
 return 0;

This is the compiler error

lmbfwd.cpp: In instantiation of ‘enqueue(T&&, Args&& ...)::<lambda()> [with T = main()::<lambda(int, int)>&; Args = int, int]’:
lmbfwd.cpp:11:27: required from ‘struct enqueue(T&&, Args&& ...) [with T = main()::<lambda(int, int)>&; Args = int, int]::<lambda()>’
lmbfwd.cpp:10:2: required from ‘void enqueue(T&&, Args&& ...) [with T = main()::<lambda(int, int)>&; Args = int, int]’
lmbfwd.cpp:18:20: required from here
lmbfwd.cpp:11:26: error: no matching function for call to ‘forward<int>(const int&)’
 func(std::forward<Args>(args)...);

I am not able to understand why argument forwarding fails without mutable.

Besides, if I pass a lambda with string as argument, mutable is not required and program works.

#include <iostream>
#include <queue>
#include <functional>

std::queue<std::function<void()>> q;

template<typename T, typename... Args>
void enqueue(T&& func, Args&&... args)

 //works without mutable
 q.emplace([=]() 
 func(std::forward<Args>(args)...);
 );

void dequeue()

 while (!q.empty()) 
 auto f = std::move(q.front());
 q.pop();
 f();
 

int main()

 auto f3 = [](std::string s) std::cout << s << "n"; ;
 enqueue(f3, "Hello");
 dequeue();
 return 0;

Why is mutable required in case of int double and not in case of string ? What is the difference between these two ?

A non-mutable lambda generates a closure type with an implicit const qualifier on its operator() overload.

std::forward is a conditional move: it is equivalent to std::move when the provided template argument is not an lvalue reference. It is defined as follows:

template< class T >
constexpr T&& forward( typename std::remove_reference<T>::type& t ) noexcept;

template< class T >
constexpr T&& forward( typename std::remove_reference<T>::type&& t ) noexcept;

(See: https://en.cppreference.com/w/cpp/utility/forward).

Let's simplify your snippet to:

#include <utility>

template <typename T, typename... Args>
void enqueue(T&& func, Args&&... args)

 [=] func(std::forward<Args>(args)...); ;


int main()

 enqueue([](int) , 10);

The error produced by clang++ 8.x is:

error: no matching function for call to 'forward'
 [=] func(std::forward<Args>(args)...); ;
 ^~~~~~~~~~~~~~~~~~
note: in instantiation of function template specialization 'enqueue<(lambda at wtf.cpp:11:13), int>' requested here
 enqueue([](int) , 10);
 ^
note: candidate function template not viable: 1st argument ('const int')
 would lose const qualifier
 forward(typename std::remove_reference<_Tp>::type& __t) noexcept
 ^
note: candidate function template not viable: 1st argument ('const int')
 would lose const qualifier
 forward(typename std::remove_reference<_Tp>::type&& __t) noexcept
 ^

In the snippet above:

• Args is int and refers to the type outside of the lambda.




• args refers to the member of the closure synthesized via lambda capture, and is const due to the lack of mutable.

Therefore the std::forward invocation is...

std::forward<int>(/* `const int&` member of closure */)

...which doesn't match any existing overload of std::forward. There is a mismatch between the template argument provided to forward and its function argument type.

Adding mutable to the lambda makes args non-const, and a suitable forward overload is found (the first one, which moves its argument).

By using C++20 pack-expansion captures to "rewrite" the name of args, we can avoid the mismatch mentioned above, making the code compile even without mutable:

template <typename T, typename... Args>
void enqueue(T&& func, Args&&... args)

 [func, ...xs = args] func(std::forward<decltype(xs)>(xs)...); ;

live example on godbolt.org

Why is mutable required in case of int double and not in case of string ? What is the difference between these two ?

This is a fun one - it works because you're not actually passing a std::string in your invocation:

enqueue(f3, "Hello");
// ^~~~~~~
// const char*

If you correctly match the type of the argument passed to enqueue to the one accepted by f3, it will stop working as expected (unless you use mutable or C++20 features):

enqueue(f3, std::string"Hello");
// Compile-time error.

To explain why the version with const char* works, let's again look at a simplified example:

template <typename T>
void enqueue(T&& func, const char (&arg)[6])

 [=] func(std::forward<const char*>(arg)); ;


int main()

 enqueue([](std::string) , "Hello");

Args is deduced as const char(&)[6]. There is a matching forward overload:

template< class T >
constexpr T&& forward( typename std::remove_reference<T>::type&& t ) noexcept;

After substitution:

template< class T >
constexpr const char*&& forward( const char*&& t ) noexcept;

This simply returns t, which is then used to construct the std::string.