CODEWITHSUNDEEP
c++lambdac++14auto
Kumar Roshan Mehta
Kumar Roshan Mehta

Reputation: 3306

Simple lambda function I could not understand

I am learning C++14 lambdas with const, and today my friend showed me the following. I could not understand it.

  1. Is it a lambda function? The syntax does not matches what I see usually.

  2. it syntax matches with a lambda function, but it fails with a long error.

int main()
{
    // 1.
    const auto x = [&]{
        auto l = 0;
        l = 99;
        return l;
    }();

    std::cout << x << endl;

    // 2.    
    const auto y = [&](){
        auto l = 0;
        l = 99;
        return l;
    };

    std::cout << y << endl;   

    return 0;
}

I want to know what 1 is, and why 2 fails to compile.

Upvotes: 6

Views: 349

Answers (1)

max66
max66

Reputation: 66200

I wanted to know what is 1. and why 2. fails to compile.

(1)

const auto x = [&]{
        auto const_val = 0;
        const_val = 99;
        return const_val;
    }();
// ..^^  <--- execution

This is the definition and execution of a lambda that doesn't receive arguments (so the () part after [&] is optional and, in this case, omitted).

So x is an int (a const int) initialized with 99 (the value returned by the lambda)

As you can see, the name const_val for the integer variable inside the lambda is a fake, because the variable is intialized with 0 and then modified assigning to it the value 99.

(2)

const auto y = [&](){
auto l = 0;
l = 99;
return l;
};

This is the definition only (no execution) of a lambda that receive no arguments.

So y is a variable (well, a constant) that contain the lambda and when you write

std::cout << y << endl;

you get an error because isn't defined the output for a lambda; you should try with

std::cout << y() << endl;

to execute the lambda and print the returned value (again 99).

Upvotes: 16

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