Troskyvs Troskyvs - 1 year ago 38
C++ Question

C++11: The mutable lambda doesn't seem to change the variable?

I've got a quick test below:

#include<iostream>
using namespace std;
int main(){
int i=2;
auto f=[=]()mutable{++i;};
f();
f();
cout<<i<<endl;
return 0;
}


But the result it still prints "2". Why i is not modified inside a mutable lambda? I'm using clang --std=c++1z.

Thanks!

Answer Source
int i=2;
auto f=[=]()mutable{++i;};
f();
f();
std::cout<<i<<std::endl;

this prints 2.

int i=2;
auto f=[&](){++i;};
f();
f();
std::cout<<i<<std::endl;

this prints 4.

int i=2;
auto f=[=]()mutable{++i; std::cout << i << std::endl;};
f();
f();
std::cout<<i<<std::endl;

this prints 3 4 2.

= copies captured data into the lambda.

If mutable the copies can be modified.

& references captured data in the lambda.

Modifying things through references is legal.

[=] is the same as [i], and [&] is the same as [&i] in this context (you can explicitly list captures, or let them be captured implicitly by listing none and using = or &).

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