char * x，y，z; char * x，y，z; char（*）x，y，z ;?之间有什么区别？

Question

Okay I am getting lost in these pointers can somebody exactly tell me what is(are) the difference between char * x,y,z; , char* x,y,z; and char (*)x,y,z; ? If you can please provide resources to your answer or something.

Answer 1

The first two mean the same thing. They declare x as a pointer to a char, and y and z as char variables. The third one will cause a syntax error and, as @danielfranca pointed out in the comments, will not compile.

Answer 2

These two records

char * x,y,z; 
char* x,y,z;

are identical and equivalent to

char *x;
char y;
char z;

Take into account that these declarations are equivalent

char*x;
char* x;
char * x;
char *x;

All them declare variable x as pointer to char.

This record

char (*)x,y,z;

is invalid and will not be compiled.

I think you mean the following

char (*x),y,z;

In this case declaration

char ( *x );

is equivalent to

char *x;

You may enclose in parentheses a declarator. So the above record you could write like

char ( *x ), ( y ), ( z );

Answer 3

If you are asking the question, it is probably because you were looking for a shorthand way of declaring x, y and z all to be pointers to characters.

In most cases, you should just be clear about it:

char *x, *y, *z;

In more complicated situations where you will use it often, you can use a typedef:

typedef char *cp_t;
cp_t x, y, z;

Answer 4

Declare variables in the same way as you're gonna use them.

char *x, y, z;

*x is a char, y is a char, z is a char. So x is a pointer to a char.

void f(void), (*g)(void);

f and *g are functions without parameters or return value. So g is a pointer to such a function. The parentheses around *g are needed because (void) binds more tightly than * .