What happens when I declare say mul开发者_C百科tiple 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?
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.
In your first sentence:
int x, y, z;
They are all int
s.
However, in the second one:
int* x, y, z;
Only x
is a pointer to int
. y
and z
are plain int
s.
If you want them all to be pointers to int
s you need to do:
int *x, *y, *z;
Only x is an int pointer. Y and Z will be just int. If you want three pointers:
int * x, * y, * z;
It is important to know that, in C, declaration mimics usage. The * unary operator is right associative in C. So, for example in int *x
x is of the type pointer to an int (or int-star) and in int x
, x is of type int.
As others have also mentioned, in int* x, y, z;
the C compiler declares x as an int-star and, y and z as integer.
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