CODEWITHSUNDEEP
c++reference
newprogrammer
newprogrammer

Reputation: 2604

Is there a difference between int& a and int &a?

This code:

int a = 5;
int& b = a;
b = 7;
cout << a;

prints out 7, and replacing int& b with int &b also prints out 7.

In fact so does int&b and int & b.

I tested this kind of behavior with a simple class as well. In general, does it ever matter whether the ampersand is placed relative to the type and identifier? Thanks.

Upvotes: 17

Views: 19581

Answers (6)

Agaz Wani
Agaz Wani

Reputation: 5694

These are two functionally equivalent declarations:

int& a; // & associated with type
int &a; // & associated with variable

Associating the & or * with the type name reflects the desire of the programmer to have a separate pointer type. However, the difficulty of associating the & or * with the type name rather than the variable is that, according to the formal C++ syntax, neither the & nor the * is distributive over a list of variables. Thus, easily creating misleading declarations. For example, the following declaration creates one, not two, integer references.

int& a, b;

Here, b is declared as an integer (not an integer reference) because, when used in a declaration, the & (or *) is linked to the individual variable that it precedes, not to the type that it follows. The problem with this declaration is that, visually both a and b seem to be of reference types, even though, only a is a reference, thus visual confusion not only misleads novice C++ programmers, but occasionally experienced programmers too.

It doesn't matter whether you write int &a or int& a for a C++ compiler. However, to avoid confusion, & and * should be associated with variable rather than type.

Upvotes: 3

cammando
cammando

Reputation: 616

It doesn't make any difference because c/c++/java these languages are not space sensitive.

Upvotes: 1

Pubby
Pubby

Reputation: 53097

In general, does it ever matter whether the ampersand is placed relative to the type and identifier?

The example posted it does not matter, but there are cases where it does:

int & a, b;    // b is not a ref
int const * a; // pointer to const int
int * const a; // const pointer to int

As always, whitespace doesn't matter.

Upvotes: 7

Ben Voigt
Ben Voigt

Reputation: 283961

In C-like languages, whitespace mostly doesn't matter.

All versions you listed parse as the same three tokens:

  • int
  • &
  • b

so they mean the same to the compiler.

The only time whitespace matters is when it separates two alphanumeric tokens, and even then the amount and type of whitespace doesn't matter, as long as there is some. But any sort of punctuation always becomes a separate token from alphanumerics, no whitespace is needed.

So your code can become as short as:

using namespace std;int a=5;int&b=a;b=7;cout<<a;

The remaining whitespace is needed.

Upvotes: 11

vdbuilder
vdbuilder

Reputation: 12994

It doesn't make a difference to the compiler which way it is written. Which is correct is a matter of code style.

Upvotes: 1

Ry-
Ry-

Reputation: 225281

No, there is absolutely no difference except coding style. I think the main argument about coding style is that this declaration:

int& a, b;

declares a as an int& and b as an int.

Upvotes: 21

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