Suppose if I write the following code
int i=10; int &j=i; //a reference in C++,don't confuse it with pointers & address Does j takes any space in the memory as its simply a reference?
3 Answers
You can't bind a non-const reference to a const value. The compiler should give you an error:
invalid initialization of reference of type ‘int&’ from expression of type ‘const int’
Even if you do get around this (with, for example, a const_cast), the code will have undefined behavior because you're modifying an originally const value.
Does i takes any space in the memory as its simply a reference?
That's an implementation detail - it could be optimized out completely. What you need to know is that j is just a nickname for i.
Comments
First of all you are doing invalid initialization of &j as said by Luchian.
A reference is just a name given to the variable's memory location.You can access the contents of the variable through either the original variable name or the reference. You can read the following declaration :
int &j=i; as:
j is an integer reference initialized to i.
Well, their occupying space is implementation dependent,Its not that they are stored somewhere, you can think of them just as a label.
From C++ Standard (§ 8.3.2 #4)
It is unspecified whether or not a reference requires storage
The emphasis is mine.
*NOTE--*References are usually used for function argument lists and function return values.
2 Comments
Nikos C.
Compilers implement references as pointers and thus they do consume space.
James Kanze
@NikosC. Sometimes. As a local variable, I'd be surprised if any compiler allocated memory for the reference.
Error when you try to assign the reference to a constant value "error C2440: 'initializing' : cannot convert from 'const int' to 'int &'"
int &j=i;is invalid conversion.jis a reference ,if you are confusing it with pointers.