C++ Lambdas: Difference between “mutable” and capture-by-reference

Question

In C++ you can declare lambdas for example like this:

int x = 5;
auto a = [=]() mutable { ++x; std::cout << x << \'\\n\'; };
auto b = [&]()               


        

Answer 1

What is happening

The first will only modify its own copy of x and leave the outside x unchanged. The second will modify the outside x.

Add a print statement after trying each:

a();
std::cout << x << "----\n";
b();
std::cout << x << '\n';

This is expected to print:

6
5
----
6
6

Why

It may help to consider that lambda

[...] expressions provide a concise way to create simple function objects

^{(see [expr.prim.lambda] of the Standard)}

They have

[...] a public inline function call operator [...]

which is declared as a const member function, but only

[...] if and only if the lambda expression’s parameter-declaration-clause is not followed by mutable

You can think of as if

    int x = 5;
    auto a = [=]() mutable { ++x; std::cout << x << '\n'; };

==>

    int x = 5;

    class __lambda_a {
        int x;
    public:
        __lambda_a () : x($lookup-one-outer$::x) {}
        inline void operator() { ++x; std::cout << x << '\n'; }     
    } a;

and

    auto b = [&]()         { ++x; std::cout << x << '\n'; };

==>

    int x = 5;

    class __lambda_b {
        int &x;
    public:
        __lambda_b() : x($lookup-one-outer$::x) {}
        inline void operator() const { ++x; std::cout << x << '\n'; }         
        //                     ^^^^^
    } b;

Q: But if it is a const function, why can I still change x?

A: You are only changing the outside x. The lambda's own x is a reference, and the operation ++x does not modify the reference, but the refered value.

This works because in C++, the constness of a pointer/reference does not change the constness of the pointee/referencee seen through it.