Call by reference C++ confusion

[shivajikobardan]

Homework Statement

Why do ostream &os? What does it do? Why not just do "ostream os"? What happens in memory when we do this call by reference thing?

Relevant Equations

none

Full code goes here: The goal of the program is to overload the extraction operator for cout.

#include<iostream>
using namespace std;class Time
{
private:
    int hour;
    int min;
public:
    Time()
    {
        hour=min=0;
    }
    Time(int h,int m)
    {

        hour=h;
        min=m;
    }
    Time operator+(Time t)
    {
        Time t1;
        int m=min+t.min;
        t1.hour=hour+t.hour+m/60;
        t1.min=m%60;
        return t1;

    }
    Time operator+(int x)
    {

        Time t;
        t.hour=hour+x;
        t.min=min;
        return t;
    }
    friend void operator<<(ostream &os,Time t );
};

void operator<<(ostream &os,Time t)
{

    cout<<"Hour : "<<t.hour<<"min:"<<t.min<<endl;
}int main()
{
    Time t1;
    Time t2(8,40);
    Time t3(5,30);
    Time t4(2,30);
    t1=t2+t3+t4;
    cout<<t1;
    return 0;
}

Please guide how to understand this. Tell me any prerequisite that I'm missing, that'd be immensely helpful than giving me hand holded solution.

 

[shivajikobardan]

When I do just "ostream os" instead of "ostream &os", I get the below error;

If anyone could share figures for what's happening in memory, that'll be hugely useful. A picture is worth 1000 words! :)

 

[Mark44]

The best explanation I can give, which might not be 100% accurate, is that the overloaded << operator expects its first argument to be a reference to the ostream class, not the class itself. This restriction is the same as the ostream class member operator.

 

[shivajikobardan]

thank you for the reply. I need to learn about call by address and call by reference first.

 

[Mark44]

thank you for the reply. I need to learn about call by address and call by reference first.

It's somewhat confusing, IMO. In the time before C++, the two types of calls were call by value and call by reference. These call types refer to what happens with a parameter when a function is called or invoked. In call by value, the function receives the value of the parameter, so if a function modifies that parameter, the change isn't reflected in the caller's variable, In call by reference, the function actually has access to the parameter, so any change that the function makes to the parameter is reflected in the caller's variable. In C, all calls were by value, but call by reference was simulated by making the parameter an address; i.e., a pointer to a variable. Because the pointer was passed by value, no the pointer's value doesn't change, but what's at that address can change.

C++ added a new type of parameter, a reference parameter. It works almost identically to the way that call by pointer address works, except that the semantics are different. IOW, a call by pointer address looks different from a call with a reference parameter.

Here's a short example program that shows all three function call types.

#include<iostream>
using std::cout; using std::endl;

void CallByValue(int val)
{
	val = 2;
}

void CallByPtr(int* pVal)
{
	*pVal = 2;
}

void CallByRef(int& val)
{
	val = 2;
}int main()
{
	int num = 5;
	// Call by value can't change value of its argument.
	CallByValue(num);
	cout << "num = " << num << endl;

	// Call by pointer address can't change value of its argument,
	//   but can change the value pointed to.
	CallByPtr(&num);
	cout << "num = " << num << endl;

	num = 5;        // Reset num.
	// Call by reference can change the value of its argument.
	CallByRef(num);
	cout << "num = " << num << endl;	
}

Note that only CallByRef() can modify its argument. CallByPtr() doesn't modify its argument, &num, but it modifies what's at that address, namely, num. CallByValue() can't modify its argument at all.

It's interesting to me that the assembly code that MSFT Visual Studio generates for CallByPtr() and CallByRef() are exactly the same, even though the way these functions are called is different. Of course, the code that the compiler generates is implementation-dependent, so other compilers might do something different.

 

[shivajikobardan]

why's your name written as Chatgpt? what happened? are you copying answers from chatgpt?

 

[phinds]

why's your name written as Chatgpt? what happened? are you copying answers from chatgpt?

What day is it?

 

[Mark44]

why's your name written as Chatgpt?

Not sure, but I think that it's a prank played by the forum's owner on all the mentors as an April Fools' Day prank.

What I wrote is NOT from ChatGPT...

 

[phinds]

an April Fools' Day prank.

Yah think ?