CODEWITHSUNDEEP
c++functionreference
sarath
sarath

Reputation: 543

C++ reference and lvalue

Can anybody please tell me what's happening in the following code. I was expecting lvalue error and it will happen if I remove the reference return type of the function. But it's giving me output as 20. Please explain. Thanks.

int &fun()
{
    static int x = 10;
    return x;
}
int main()
{
    fun() = 20;
    cout << fun();
    return 0;
}

Upvotes: 0

Views: 108

Answers (2)

Jeff
Jeff

Reputation: 3525

Basic C++ semantics:

  • You can't assign to the return values of functions unless the return type is a reference.
  • A function can't return a reference to a local variable unless it's a static variable.
  • Static variables' values persist between accesses.

If you want to assign something to x outside of fun(), x has to live somewhere, right? Making it static gives it a permanent spot that will be re-accessed every time. That's why the value of 20 persists.

Upvotes: 2

Rakib
Rakib

Reputation: 7625

The output is as expected. In the call fun() = 20;, the actual x in fun() is assigned 20, since fun() returns reference to x. In the call cout<<fun();, the assigned value is printed, which is 20.

since x is declared static, it's available in memory even after func() returns. A method scoped static variable is created when first encounter in the method, and retains until program terminates. static variable is initialized only once, and subsequent method call will see the last updated value. more on static here and here.

Upvotes: 3

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