Function pointers

Function pointers

[kumar11]

Could you please explain the meaning of below declaration statement.

char (*pa())[4]; 


Please find the entire code below.

char (*pa())[4]; 
void main(){ 
char(*p)[4]=pa(); 
printf("%d",**p); 
} 
char (*pa())[4]{ 
static char arr[]={'\11','\12','\13','\14'}; 
return &arr; 
}
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Re: Function pointers

[saravanaraj v]

[root@build ~]# cdecl
Type `help' or `?' for help
cdecl> explain char (*pa())[4];
declare pa as function returning pointer to array 4 of char
cdecl> explain char (*p)[4]
declare p as pointer to array 4 of char
cdecl>


Try the following code.

#include <stdio.h>

char (*pa())[4];
int main(){
        char(*p)[4]=pa();
        /*printf("%d",**p);*/
        printf("%d %d %d %d\n",(*p)[0],(*p)[1],(*p)[2],(*p)[3]);
        return 0;
}
char (*pa())[4]{
        static char arr[]={'\11','\12','\13','\14'};
        return &arr;
}


On Thu, Jan 21, 2010 at 12:03 PM, kumar11 <ravikumar.chakram@gmail.com> wrote:
>
> Could you please explain the meaning of below declaration statement.
>
> char (*pa())[4];
>
>
> Please find the entire code below.
>
> char (*pa())[4];
> void main(){
> char(*p)[4]=pa();
> printf("%d",**p);
> }
> char (*pa())[4]{
> static char arr[]={'\11','\12','\13','\14'};
> return &arr;
> }
> --
> View this message in context: http://old.nabble.com/Function-pointers-tp27253432p27253432.html
> Sent from the linux-c-programming mailing list archive at Nabble.com.
>
> --
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Re: Function pointers

[Glynn Clements]

kumar11 wrote:

> Could you please explain the meaning of below declaration statement.
> 
> char (*pa())[4]; 

"pa" is a function returning a pointer to an array of four characters.

C declarations work by declaring an expression involving the
identifier as having a named type (in this case, "char").

"(*pa())[i]" is a character, for 0 <= i < 4.

Therefore:

"(*pa())" is an array of four characters.

Therefore:

"pa()" is a pointer to an array of four characters.

Therefore:

"pa" is a function returning a pointer to an array of four characters.

-- 
Glynn Clements <glynn@gclements.plus.com>

Re: Function pointers

[Milind A Choudhary]

On Thu, Jan 21, 2010 at 4:56 AM, Glynn Clements
<glynn@gclements.plus.com> wrote:
>
> kumar11 wrote:
>
>> Could you please explain the meaning of below declaration statement.
>>
>> char (*pa())[4];
>
> "pa" is a function returning a pointer to an array of four characters.
>

You already have got the required answer from others.
C and function pointers can always be confusing. At times you think
you have understood it throughly and but still not be able to decode some
more complex declarations. (cdec can come to help but not always..at least not
in a technical interview ) The best approach that I have come across is
the `Clockwise Spiral` rule. (link :
http://www.c-faq.com/decl/spiral.anderson.html )

Hope this helps..

Thanks & Regards
Milind A Choudhary
http://www.csl.mtu.edu/~machoudh/

-- 
Milind  Arun Choudhary

Re: Re: Re: Function pointers

[Glynn Clements]

simon wrote:

> char (*pa())[4]; 
> is this mean that an pa function pointer with return value like a char a[4] and it's argument is nothing ?
> 
> right?

Wrong.

1. pa is not a pointer to a function, it's a function which returns a
pointer. "()" has higher precedence than "*", so "*pa()" is parsed as
"*(pa())" (function returning pointer) rather than "(*pa)()" (pointer
to function).

2. An empty parameter list "()" indicates that the function takes
unspecified arguments. A function which takes no arguments has
"(void)" as the parameter list.

-- 
Glynn Clements <glynn@gclements.plus.com>