question about pointer define

[Wonder]

Hello,

I'm confused by the pointer definition such as int *(p[3]);

It seems if the parenthesis close p[3], it defines only 3 integers. The
star
is just useless. It can be showed by my program:

int main()
{
int *(p[3]);
// cout << *p[0] << "\t" << *p[1] << "\t" << *p[2] << endl;
cout << &p[0] << "\t" << &p[1] << "\t" << &p[2] << endl;
cout << p[0] << "\t" << p[1] << "\t" << p[2] << endl;
}

The program can be compiled on cygwin linux by g++ and run, the output
is:

0x22efb0 0x22efb4 0x22efb8
0x76c 0xffffffff 0x22efc8

If you delete the * in the definition, define int (p[3]); the second
line
will be:

1900 -1 2289608

Which is exactly the decimal numbers of the the previous program.

The second line is comment-out, coz it'll be core dumped if you want to
get
the content of the pointer.

Could someone explain what's going on here? Thanks.

[Emmanuel Delahaye]

Wonder wrote on 08/09/05 :

// cout << *p[0] << "\t" << *p[1] << "\t" << *p[2] << endl;
cout << &p[0] << "\t" << &p[1] << "\t" << &p[2] << endl;
cout << p[0] << "\t" << p[1] << "\t" << p[2] << endl;

Not a C-code. C++ is next door.

--
Emmanuel
The C-FAQ: http://www.eskimo.com/~scs/C-faq/faq.html
The C-library: http://www.dinkumware.com/refxc.html

"There are 10 types of people in the world today;
those that understand binary, and those that dont."

[Wonder]

i'm just using cout to output. It has nothing to do with the nature of
the question here.
I'm new here, don't know we have to write strict C code.

[Irrwahn Grausewitz]

"Wonder" <sa******@gmail .com> wrote:

I'm confused by the pointer definition such as int *(p[3]);

It seems if the parenthesis close p[3], it defines only 3 integers.
No. p is an array of three pointers to int.
The star is just useless.
Definitely not. The parentheses are.
It can be showed by my program:

int main()
{
int *(p[3]);
// cout << *p[0] << "\t" << *p[1] << "\t" << *p[2] << endl;

<snip>

This is by no means a C program. Either rewrite in C, or take this to
comp.lang.c++, please.

Best regards
--
Irrwahn Grausewitz (ir*******@free net.de)
welcome to clc : http://www.ungerhu.com/jxh/clc.welcome.txt
clc faq-list : http://www.faqs.org/faqs/C-faq/faq/
clc frequent answers: http://benpfaff.org/writings/clc.

[Wonder]

i don't think the elements in p are pointers, coz you even can't access
*p[i], i=0,1,2

if you make the parentheses close *p, or just erase parentheses, such
as
int (*p)[3] or int *p[3]

no problem, it's the definition of an array of three pointers to int.

BTW: is there a rule say that only C program can be written here, even
though the question is about C in nature?

Thanks.

[John Bode]

Wonder wrote:

Hello,

I'm confused by the pointer definition such as int *(p[3]);

It's the same as

int *p[3];

i.e., a 3-element array of pointers to int.
It seems if the parenthesis close p[3], it defines only 3 integers. The
star
is just useless.
The '*' indicates that p is an array of pointers to int, instead of
just an array of int.
It can be showed by my program:

int main()
{
int *(p[3]);
// cout << *p[0] << "\t" << *p[1] << "\t" << *p[2] << endl;
cout << &p[0] << "\t" << &p[1] << "\t" << &p[2] << endl;
cout << p[0] << "\t" << p[1] << "\t" << p[2] << endl;
}

The program can be compiled on cygwin linux by g++ and run, the output
is:

0x22efb0 0x22efb4 0x22efb8
0x76c 0xffffffff 0x22efc8

If you delete the * in the definition, define int (p[3]); the second
line
will be:

1900 -1 2289608

Which is exactly the decimal numbers of the the previous program.

Right. What's happening is that the contents of p are uninitialized,
so p[0] through p[2] contain random bit strings, and these random bit
strings happened to be the same between the two runs. In the first
case, where p is declared as

int *(p[3]);

each element of p is of a *pointer* type, so the output is formatted to
reflect that (i.e., as hex values corresponding to a virtual memory
address). In the second case, where p is declared as

int (p[3]);

each element is of an *int* type, and the output is formatted to
reflect that.
The second line is comment-out, coz it'll be core dumped if you want to
get
the content of the pointer.

Right, because without proper initialization, p[0] through p[2] don't
point anywhere meaningful.
Could someone explain what's going on here? Thanks.

Basically, for any type T and declarator foo,

T *(foo);

is the same as

T *foo;

[Martin Ambuhl]

Wonder wrote:

Hello,

I'm confused by the pointer definition such as int *(p[3]);

It seems if the parenthesis close p[3], it defines only 3 integers.
Wrong.
The star is just useless.
Wrong.
It can be showed by my program:
You program shows nothing. It is nothing but a bunch of syntax errors.
If you had posted it to <news:comp.lang .c++> instead, it would still be
garbage: no inclusion of needed the header file, no specification of the
needed namespace qualification.

int main()
{
int *(p[3]);
// cout << *p[0] << "\t" << *p[1] << "\t" << *p[2] << endl;
cout << &p[0] << "\t" << &p[1] << "\t" << &p[2] << endl;
cout << p[0] << "\t" << p[1] << "\t" << p[2] << endl;
}

The program can be compiled on cygwin linux by g++
Which is not a C compiler, therefore not topical in <news:comp.lang .c>.
And it won't compile under g++ if you invoke g++ with appropriate
specification of diagnostics and of the standard used.

[...] Could someone explain what's going on here? Thanks.

You're an idiot.

[Martin Ambuhl]

Wonder wrote:

i'm just using cout to output. It has nothing to do with the nature of
the question here.
I'm new here, don't know we have to write strict C code.

Your code is not C. It is not loose C; it is not strict C; it is not
GNU C. It is in a different language.

[Wonder]

Thanks a lot for your kind reply.

Now the problem goes to the difference between
int (*p)[3];
int *p[3];
int *(p[3]);

As you said, the second and the third are same. You are right, for
both, the *p[i] will give a core dump. However, the first one just give
the values of the elements, even though I didn't initialize anything.

int main()
{
int (*p)[3];
cout << *p[0] << "\t" << *p[1] << "\t" << *p[2] << endl;
cout << &p[0] << "\t" << &p[1] << "\t" << &p[2] << endl;
cout << p[0] << "\t" << p[1] << "\t" << p[2] << endl;
}
The output is:
440 -1869574000 -1869574000
0x6101cf3e 0x6101cf4a 0x6101cf56
0x6101cf3e 0x6101cf4a 0x6101cf56

Another strange thing is the second line and third line are same in
this case.

[Wonder]

I didn't copy & paste the header coz I thought it had nothing to do
with the question itself, and I don't want to waste too many lines.

You just show the rudeness which should not represent the spirit of
comp.lang.c.