CODEWITHSUNDEEP
c++reference
user3557062
user3557062

Reputation: 1

difference between variable and reference assignment

This has bothered me, I have just gotten back into c++ lately, and am well aware of references. Why does it seem there is no difference between b and c below. In other words, intuitively I would think c not being a ref variable would take a copy. It got me thinking when considering the problem of returning a reference to a local variable.

int a = 10;
int& func()
{
   return a;
}

int &b = func();  
int c = func(); // shouldn't this make a copy into c but c acts like a reference.

Upvotes: 0

Views: 51

Answers (2)

user3557062
user3557062

Reputation: 1

Thanks for the answers, makes sense now. This question started from "why can't I return a local variable reference to a variable if its a copy?" Well, I realized because its too late at that point - the stack is gone. I was thinking the copy could be made before the stack went away - wrong! You can return to a reference but when you use it, segfault. Return to a variable and, segfault immediately since it tries to copy.

int& func()   // compile warning
{
   int a = 10;
   return a;
}

int &b = func();  // ok, but segfault if try to use b.
int c = func(); // segfault immediately

Upvotes: 0

user12002570
user12002570

Reputation: 1

I would think c not being a ref variable would take a copy.

c is indeed a copy of a. Note that int c = func(); is copy initialization while int &b = func(); is reference initialization.

This means that if you make changes to c they will not be reflected on a. Demo

int a = 10;
int& func()
{
   return a;
}

int &b = func();  //reference initialization
int c = func();   //copy initialization
int main()
{
    c = 20;
    std::cout << a<< " "<< b <<" "<< c <<std::endl; // 10 10 20
    return 0;
}

Note also the emphasis on "initialization" instead of "assignment". These terms have different meaning in C++ unlike other languages(python for instance).

Upvotes: 1

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