Ampersand usage in C++ ?

[LudaLuda]

Guys I got a question,

When you use the Ambersand sign "&" infront of a variable it means we are passing the address of that variable , right?

what happents when you have the "&" at the end of the variable .
Ex :
if I have a function header

exam(const K&, const V&);

what is the usgae of & in the above?

and what is the difference between this one
exam(const &K, const &V);

[bluesource]

When you use the Ambersand sign "&" infront of a variable it means we are passing the address of that variable , right?

Correct,

int x=10;
int* y=&x;

In this example we are passing the address of x to the pointer y.

what happents when you have the "&" at the end of the variable .

This refers to passing by reference.

int i=1;
int& r=i;  // r and i now refer to the same int
int x=r;   // x=1
r=2;       // i=2

*Note, references must be initialized. Also important, the reference type and variable type should NOT differ, otherwise c++ will create a hidden object that does not alias the specified value.

int value;
float& alias=value;

Very bad, the types differ.

int value;
int& alias;

Also bad, alias is not initalized.

There are reasons for using references, one is that they use a cleaner syntax than passing pointers.

When you pass a reference to a function, you eliminate copying. So if you have a very large object and don't want to modify it in the function, passing it as a

const T&

is a good idea.

[Edmund Zheng]

The first statement

exam(const K&, const V&);

is a valid funtion declartion if K and V are types. This function's parameters are refenence.

However the second statement

exam(const &K, const &V);

I do not understand. const is a key word and if K and V are types then why & is positioned before a type£¿ It is illegal against c++ grammer.

[bluesource]

exam(const &K, const &V);

I do not understand. const is a key word and if K and V are types then why & is positioned before a type£¿ It is illegal against c++ grammer.

That is a good point, I was confused by this part of the question.
The OP was confused, in the 1st posting Luda says:

what happents when you have the "&" at the end of the variable .

when Luda probably meant: "what happens when you have the "&" at the end of the type."


Likewise, Luda probably meant

exam(const K &value, const V &value2);

which is the same thing as:

exam(const K& value, const V& value2);

So to answer the final question, what is the usage of exam(const K&, const V&); ?

Take the following example:

exam(const K&, const v&);

K a=10;
K& b=a;

V c=11;
V& d=c;

exam(b,d);

This illustrates passing by reference, your function exam will modify the a and c variables by reference, even though you passed b and d in.

[bluesource]

I hope this wasn't overly confusing...but now after I've thought about it I think your confusion can be summarized as follows:

int x=10;
int& y=x;

We say y references x.

But this is saying the same thing as

int x=10;
int &y=x;

Once again we say y references x even though they were declared differently. The & does not refer to the address.
So

int x=10;
int* y=&x;

Here the & refers to the address, whereas in the above two examples it does NOT.

[LudaLuda]

Thanks guys!

I guess my question was a lil confusing. sorry.

but Im all clear now.

[gstercken]

Originally posted by LudaLuda
I guess my question was a lil confusing. sorry.

Don't worry, the answers have been not less confusing...