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 ?
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