return by reference : what's different here ??

[sathya691]

Hi guys,

i've two functions declared like the following.

int& readRef(int& num){
return num;
}

int readNormal(int& num){
return num;
}

I thought the function readRef() returns a reference to num, so i can change the value of through that reference. But, it's not

happening. Effectively, there's no difference between these two functions, then. But, it can't be. I would be very thankful to the

guy(s) who explain where i was wrong.

[mtrenkmann]

The thing is that you pass the int by value in both cases. Then it doesn't matter if you return this local copy of num by value or by reference (your compiler should complain about this version) ...

Here is what you probably want:

int& readRef(int& num) {
  // do something
  return num;
}

[sathya691]

@mtrenkmann : thanks. it's a typing mistake. I actually meant it the way you said. But, even then it's not performing as i thought. Can you help me out?

[therockon7throw]

Hi,
You cant send a reference to a local variable(member), however if you still need it, you may do some thing like following

int& readRef(int* i)
{
	return *(i+0);
}

and you may change the value of the variable( member) like following

readRef( &t) = 18;

assuming that t is a type of int

[sathya691]

@therockon7throw : Thank you. Thank you very much.

[mtrenkmann]

@therockon7throw

You cant send a reference to a local variable(member)

Thats right, but in his/her case the variable is not local because its passed by reference. Then you can return by reference, its just kind of forwarding, no problem.

readRef( &t) = 18;

You should never write code like this. What is the intent? Does readRef change the value of its argument? And if so, this value would be immediately overridden! So why not simply writing i = 18;? Has readRef any side effects? Very confusing!

Furthermore your code is incorrect: return *(i+0); adds zero to an address, which makes no sense.

@sathya691 Can you post an example of your intended usage? You definitely should go with references instead of pointers. There are reasons why C++ has them.

[therockon7throw]

hi mtrenkmann,

you are right saying

You should never write code like this. What is the intent? Does readRef change the value of its argument? And if so, this value would be immediately overridden! So why not simply writing i = 18;? Has readRef any sideeffects? Very confusing!

sathya691 needs to change the value of the argument, not inside the function, but somewhere else by sending the reference.

If sathya691's intend was to change the value of the argument, inside the readRef, he could do what you've just said

And if so, this value would be immediately overridden! So why not simply writing i = 18;

so what is the intend of readRef( &t) = 18; the same as if you send a reference by overloading the () or [] operator in array classes.

Furthermore your code is incorrect: return *(i+0); adds zero to an address, which makes no sense.

it is not incorrect, it is the same as return *i; or the same as return i[0]; it is the way i am using for examples instead of a = ar[9 ]; i use a = *( ar + 9 ) ;

I think it is a matter of taste

[mtrenkmann]

Well, ok the code is not incorrect in terms of invalid C++ code, but it is just not readable and this is no matter of taste. High-level languages such as C++ are for readability first, for humans, otherwise we could all write assembly.

Then I don't understand why you came up with things like arrays and operator overloading. The original post has nothing to do with it. Lets focus on the problem he/she has. Thanks.

[therockon7throw]

@mtrenkmann

Then I don't understand why you came up with things like arrays and operator overloading. The original post has nothing to do with it

You asked :

readRef( &t) = 18;

You should never write code like this. What is the intent?

And that was my answer :

so what is the intend of readRef( &t) = 18; the same as if you send a reference by overloading

the () or [] operator in array classes.

A question, how do change value of an element of a class member being a dynamic array outside the class , some where else ?