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85

answers:

3

What happens when I declare say multiple variables on a single line? e.g.

int x, y, z;

All are ints. The question is what are y and z in the following statement?

int* x, y, z;

Are they all int pointers?

+3  A: 

In your first sentence:

int x, y, z;

They are all ints.

However, in the second one:

int* x, y, z;

Only x is a pointer to int. y and z are plain ints.

If you want them all to be pointers to ints you need to do:

int *x, *y, *z;
Pablo Santa Cruz
This is one reason the spacing `int* foo;` is really misguided. In this question's example, it makes it look like the 3 variables are all of type "`int*`". I would say this spacing is almost as offensive as `a+b * c`.
R..
+2  A: 

Only x is an int pointer. Y and Z will be just int. If you want three pointers:

int * x, * y, * z;
Macmade
+5  A: 

Only x is a pointer to int; y and z are regular ints.

This is one aspect of C declaration syntax that trips some people up. C uses the concept of a declarator, which introduces the name of the thing being declared along with additional type information not provided by the type specifier. In the declaration

int* x, y, z;

the declarators are *x, y, and z (it's an accident of C syntax that you can write either int* x or int *x, and this question is one of several reasons why I recommend using the second style). The int-ness of x, y, and z is specified by the type specifier int, while the pointer-ness of x is specified by the declarator *x (IOW, the expression *x has type int).

If you want all three objects to be pointers, you have two choices. You can either declare them as pointers explicitly:

int *x, *y, *z;

or you can create a typedef for an int pointer:

typedef int *iptr;
iptr x, y, z;

Just remember that when declaring a pointer, the * is part of the variable name, not the type.

John Bode
Truly a complete answer and to that of the underlying question of whether the * is for the type or the variable.
ruralcoder