CODEWITHSUNDEEP
c++
mohamed yossef
mohamed yossef

Reputation: 25

I want to understand why use reference function in this example?or The importance of reference in the function in c++?

this code, I want to understand the need to use the reference in the prototype and what will happen if it is not used?

#include<iostream>
using namespace std; 
int& fun() 
{ 
    static int x = 10; 
    return x; 
} 
int main() 
{ 
    fun() ; 
    cout<<fun(); 
    return 0; 
}

Upvotes: 0

Views: 55

Answers (2)

Klaus
Klaus

Reputation: 25623

If you get a reference, you can access the variable from the "outside" world. If you "only" have the value, you can not change the variable anymore.

int& fun() 
{   
    static int x = 10; 
    return x;  
}
int main() 
{   
    std::cout<<fun() << std::endl;
    fun()=234;
    std::cout<<fun() << std::endl;
    return 0;  
}

Upvotes: 0

463035818_is_not_an_ai
463035818_is_not_an_ai

Reputation: 122830

There is no "reference function" here. The & is part of the return type and should better be written as 

int& fun() {...

The function is returning a reference so that you can do

int main() 
{ 
    fun() = 6;
    cout<<fun();   // will print 6 
    return 0; 
}

If fun would return an int not a int& then the first line would not compile and each call to the function would return you the same value.

Note that in most cases returning a reference to a function local variable is wrong. The reason it is fine here is that x is static, meaning: it will be initialized exactly once and keeps its value across function calls.

For example this function will return a counter of how many times it got called:

int count_me() {
    static int x = 0;
    x+=1;
    return x;
}

Upvotes: 1

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