return reference

This is a discussion on return reference within the C++ Programming forums, part of the General Programming Boards category; I know how item_array[2] becomes 0 but i don't know why. since array[] is not a reference to item_array[2] how ...

  1. #1
    Registered User
    Join Date
    Mar 2003
    Posts
    3

    return reference

    I know how item_array[2] becomes 0 but i don't know why. since array[] is not a reference to item_array[2] how does the function biggest, which returns a reference to array[bigges], assign 0 to item_array[2]???

    could someone please explain this to me, thank you.

    Code:
    #include <iostream>
    
    int& biggest(int array[], int n_elements){
    
    	int index;
    	int biggest;
    	
    	biggest = 0;
    	
    	for(index = 1; index < n_elements; ++index){
    		if(array[biggest] < array[index])
    			biggest = index;
    	}
    		return(array[biggest]);
    }
    	
    int main(){
    	
    	using namespace std;
    	
    	int item_array[5] = {1, 2, 5000, 3, 4};
    	
    	
    	cout << "The biggest element is " << biggest(item_array, 5) << endl;
    	biggest(item_array, 5) = 0;
    
    	cout << item_array[2] << endl;
    	
    	return 0;
    }

  2. #2
    samurai warrior nextus's Avatar
    Join Date
    Nov 2001
    Posts
    196
    when it is returning it is returning a value..array[biggest] is the same thing as a variable name...so it is returning array[2] which contains 5000...it returns a reference to it..so when you do

    Code:
    biggest(item_array, 5) = 0;
    it sets array[2] = 0;
    remember the reason this work with your arrays because arrays are very similar to pointers....when you pass an array to a function you are actually passing the address of the array pointing to the first element so you could have done pointer notation like '*(array + 2) == array[2]'....if you do this with normal variables you are just returning a local variable which is a no-no!..
    nextus, the samurai warrior

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