CODEWITHSUNDEEP
c++function
amarVashishth
amarVashishth

Reputation: 877

Changing the value of a reference to a static variable in a function

Consider the following code snippet in C++:

#include <iostream>
using namespace std;

int &fun()
{
    static int a = 10;
    return a;
}

int main(int argc, char const *argv[])
{
    int &y = fun();
    y += 30;
    cout << fun();
    return 0;
}

Output: 40

How is output of the code snippet given above justified?

Upvotes: 1

Views: 108

Answers (2)

Barry
Barry

Reputation: 303517

fun isn't a function pointer, it's a nullary function that returns an int&. Specifically, it returns a reference to a static int named a.

So what your program does is:

int &y = fun(); // this is the first call to fun, so we initialize a.
                // it now has a value of 10, and we return a reference to it
                // so y is a reference to the static variable local to fun()

y += 30;        // y == 40
                // since y is a reference to "fun::a", a is now 40 too

cout << fun();  // a is static, so it will only get initialized once.
                // that first line is effectively a NOOP, so fun() just
                // returns a reference to it. which has a value of 40.

Upvotes: 2

nbro
nbro

Reputation: 15868

You are not using function pointers, but just storing the result of a call to fun in a reference.

a is static variable, and you are initialising a reference to that variable. A reference is just another name for a, in this case, so that's why modifying the value of the reference y, you modify also the value a, which is static, that means that its value is preserved from call to call.

Upvotes: 2

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