Function with a pointer in its declaration. Why?

Question

Not sure if the title explains it. What is the difference between:

char *s_gets(char * st, int n);

and

char s_gets(char * st, int n);

I would like to ask also why is there a pointer in the function declaration in the first place? What is its purpose?

[Explanation can be found here](http://stackoverflow.com/questions/562303/the-definitive-c-book-guide-and-list) — M.M, Sep 07 '16 at 21:01
@It still would be a pointer, wouldn't it? (But yes, that's a typical beginner's function: `void f(int **);` ... :-} — too honest for this site, Sep 07 '16 at 21:26

Mikael · Answer 1 · 2016-09-07T21:02:56.220

4

char s_gets(char * st, int n); returns a character, that's all.
char *s_gets(char * st, int n); returns a pointer to a character, which can be interpreted as a pointer to a string, if it is the programmer's will.

edited Sep 07 '16 at 21:02

answered Sep 07 '16 at 21:00

Mikael

969
12
24

@KeithThompson that's why I said it can be interpreted as one, if it is what the programmer wants. Edit: ok, got what you meant, thanks. – Mikael Sep 07 '16 at 21:03
1

Decent answer to a crap question. – Mad Physicist Sep 07 '16 at 21:09
I would reduce it to "`char` and `char*` are different types". – Eugene Sh. Sep 07 '16 at 21:11

score 1 · Answer 2 · answered Sep 07 '16 at 22:37

char *s_gets(char * st, int n);

returns a pointer to char, while

char s_gets(char * st, int n);

returns a plain char.

I would like to ask also why is there a pointer in the function declaration in the first place? What is its purpose?

It indicates that the function returns a pointer value.

C declaration syntax uses something called a declarator to convey the pointer-ness, array-ness, function-ness, or combination thereof of an item. For example, take the declaration

int *p;

The type of p is int * (pointer to int). The type is specified by the combination of the type specifier int and the declarator *p. The int-ness of p is given by the type specifier int and the pointer-ness is given by the declarator *p.

If you substitute p with f(void), you get

int *f(void);

The type of f is "function returning pointer to int". The pointer-ness and function-ness of f are specified with the declarator *f(void).

For any type T and any declarator D, the following are true:

T D;        // D is an instance of T
T D[N];     // D is an array of T
T D();      // D is a function returning T

T *D;       // D is a *pointer* to T
T *D[N];    // D is an array of pointers to T
T *D();     // D is a function returning a pointer to T

T (*D)[N];  // D is a pointer to an array of T
T (*D)();   // D is a pointer to a function returning T

In both declarations and expressions, the postfix [] subscript and () function-call operators have higher precedence than unary *, so *a[N] will always be parsed as *(a[N]) (array of pointers). To declare something as a pointer to an array or a pointer to a function, you need to explicitly group the * operator with the declarator using parentheses.

Declarators can get arbitrarily complex:

T (*D())[N];      // D is a function returning a pointer to an array of T
T (*D[N])();      // D is an array of pointers to functions returning T
T *(*(*D)[N])[M]; // D is a pointer to an array of pointers to an array of pointers to T

Function with a pointer in its declaration. Why?

2 Answers2