pointer and array

[Leon Brodskiy]

Hi,

Could please anyone clarify about pointer and array in C?

If I have:

int arr[10];

The following two commands will be the same: arr and &arr[0].

What about &arr? Is it not the same thing?

[Joona I Palaste]

Leon Brodskiy <iv**@rogers.co m> scribbled the following:

Hi, Could please anyone clarify about pointer and array in C? If I have: int arr[10]; The following two commands will be the same: arr and &arr[0].
They're expressions, not commands. But yes, they're the same thing,
in a value context.
What about &arr? Is it not the same thing?

Not the same thing at all. It's the address of the entire array, not
its first element. The most practical meaning of this is that such
addresses increment in terms of 10*sizeof(int), not in terms of
sizeof(int).

--
/-- Joona Palaste (pa*****@cc.hel sinki.fi) ------------- Finland --------\
\-------------------------------------------------------- rules! --------/

[Leon Brodskiy]

Thanks for your answer.
I'm asking because I have found in a program the following call:

void f1(char inp[10])
{...}

void main()
{...
char str[10];
....
f1(&str);
....
}

I tought this is a bug and this is not supposed to work but it does work.
Function receives pointer to char. I send to the function a pointer to a
pointer to char but still seems like in arrays it is the same pointer. So,
is this case str and &str the same?

Thanks.

"Joona I Palaste" <pa*****@cc.hel sinki.fi> wrote in message
news:cq******** **@oravannahka. helsinki.fi...

Leon Brodskiy <iv**@rogers.co m> scribbled the following:

Hi,

Could please anyone clarify about pointer and array in C?

If I have:

int arr[10];

The following two commands will be the same: arr and &arr[0].

They're expressions, not commands. But yes, they're the same thing,
in a value context.

What about &arr? Is it not the same thing?

Not the same thing at all. It's the address of the entire array, not
its first element. The most practical meaning of this is that such
addresses increment in terms of 10*sizeof(int), not in terms of
sizeof(int).

--
/-- Joona Palaste (pa*****@cc.hel sinki.fi) ------------- Finland --------\
\-------------------------------------------------------- rules! --------/

[Joona I Palaste]

Leon Brodskiy <iv**@rogers.co m> scribbled the following:

Thanks for your answer.
I'm asking because I have found in a program the following call: void f1(char inp[10])
{...} void main()
Change this to int main(void).
{...
char str[10];
...
f1(&str);
...
} I tought this is a bug and this is not supposed to work but it does work.
Function receives pointer to char. I send to the function a pointer to a
pointer to char but still seems like in arrays it is the same pointer. So,
is this case str and &str the same?

That program is *NOT* correct. You are supposed to call f1(str), not
f1(&str). If the function f1 assigns its parameter to a char pointer
(char*), it might work under your implementation. But trying to access
the char array pointer (char(*)[]) as if it were a char pointer will
yield wrong results, mainly for the reasons I stated.

--
/-- Joona Palaste (pa*****@cc.hel sinki.fi) ------------- Finland --------\
\-------------------------------------------------------- rules! --------/
"Ice cream sales somehow cause drownings: both happen in summer."
- Antti Voipio & Arto Wikla

[Joona I Palaste]

Joona I Palaste <pa*****@cc.hel sinki.fi> scribbled the following:

Leon Brodskiy <iv**@rogers.co m> scribbled the following:

Thanks for your answer.
I'm asking because I have found in a program the following call: void f1(char inp[10])
{...} void main()

Change this to int main(void). {...
char str[10];
...
f1(&str);
...
} I tought this is a bug and this is not supposed to work but it does work.
Function receives pointer to char. I send to the function a pointer to a
pointer to char but still seems like in arrays it is the same pointer. So,
is this case str and &str the same?

That program is *NOT* correct. You are supposed to call f1(str), not
f1(&str). If the function f1 assigns its parameter to a char pointer
(char*), it might work under your implementation. But trying to access
the char array pointer (char(*)[]) as if it were a char pointer will
yield wrong results, mainly for the reasons I stated.

I should not be answering C questions late in the evening. That program
is still wrong, but my answer above is wrong too. Any possible
assignment from &str to a char pointer happens *before* f1's code
begins, so what f1 does to the pointer does not matter. Does assigning
a char array pointer to a char pointer cause undefined behaviour?

--
/-- Joona Palaste (pa*****@cc.hel sinki.fi) ------------- Finland --------\
\-------------------------------------------------------- rules! --------/
"Remember: There are only three kinds of people - those who can count and those
who can't."
- Vampyra

[Leon Brodskiy]

If we leave char array pointer for a second, can we say that the program
will always work correctly?

Would it be correct to say that a difference between str and &str is that
str is char* (points to one single character) and &str is a pointer to an
array(points to array of 10 characters)? If this is a true then in the
program should not be any difference if we use &str or str - both of them
will send to the function the address of the first element in the array.

Thanks in advance.

"Joona I Palaste" <pa*****@cc.hel sinki.fi> wrote in message
news:cq******** **@oravannahka. helsinki.fi...

Joona I Palaste <pa*****@cc.hel sinki.fi> scribbled the following:

Leon Brodskiy <iv**@rogers.co m> scribbled the following:

Thanks for your answer.
I'm asking because I have found in a program the following call: void f1(char inp[10])
{...} void main()

Change this to int main(void). {...
char str[10];
...
f1(&str);
...
} I tought this is a bug and this is not supposed to work but it does work. Function receives pointer to char. I send to the function a pointer to a pointer to char but still seems like in arrays it is the same pointer. So, is this case str and &str the same?

That program is *NOT* correct. You are supposed to call f1(str), not
f1(&str). If the function f1 assigns its parameter to a char pointer
(char*), it might work under your implementation. But trying to access
the char array pointer (char(*)[]) as if it were a char pointer will
yield wrong results, mainly for the reasons I stated.

I should not be answering C questions late in the evening. That program
is still wrong, but my answer above is wrong too. Any possible
assignment from &str to a char pointer happens *before* f1's code
begins, so what f1 does to the pointer does not matter. Does assigning
a char array pointer to a char pointer cause undefined behaviour?

--
/-- Joona Palaste (pa*****@cc.hel sinki.fi) ------------- Finland --------\
\-------------------------------------------------------- rules! --------/
"Remember: There are only three kinds of people - those who can count and

those who can't."
- Vampyra

[Joona I Palaste]

Leon Brodskiy <iv**@rogers.co m> scribbled the following:

If we leave char array pointer for a second, can we say that the program
will always work correctly?
No. The C standard allows for an implementation to behave erroneously
when a pointer value is assigned to a variable of incompatible type.
Would it be correct to say that a difference between str and &str is that
str is char* (points to one single character) and &str is a pointer to an
array(points to array of 10 characters)?
Yes.
If this is a true then in the
program should not be any difference if we use &str or str - both of them
will send to the function the address of the first element in the array.

That is not true in the general case. At least I think it's not. Maybe
some of the real C gurus here can answer my original question about
whether it is undefined behaviour?

--
/-- Joona Palaste (pa*****@cc.hel sinki.fi) ------------- Finland --------\
\-------------------------------------------------------- rules! --------/
"Parthenogeneti c procreation in humans will result in the founding of a new
religion."
- John Nordberg

[Keith Thompson]

"Leon Brodskiy" <iv**@rogers.co m> writes:

If we leave char array pointer for a second, can we say that the program
will always work correctly?

Would it be correct to say that a difference between str and &str is that
str is char* (points to one single character) and &str is a pointer to an
array(points to array of 10 characters)? If this is a true then in the
program should not be any difference if we use &str or str - both of them
will send to the function the address of the first element in the array.

Please don't top-post. Your response should follow any quoted text,
not precede it.

No we can't say that the program will always work correctly. A
pointer-to-char and a pointer-to-array-of-char are two distinct types.
On most systems, they happen to have the same representation and can
be used more or less interchangeably , but the language standard
doesn't guarantee this.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Joe Wright]

Joona I Palaste wrote:

Leon Brodskiy <iv**@rogers.co m> scribbled the following:

If we leave char array pointer for a second, can we say that the program
will always work correctly?

No. The C standard allows for an implementation to behave erroneously
when a pointer value is assigned to a variable of incompatible type.

Would it be correct to say that a difference between str and &str is that
str is char* (points to one single character) and &str is a pointer to an
array(point s to array of 10 characters)?

Yes.

If this is a true then in the
program should not be any difference if we use &str or str - both of them
will send to the function the address of the first element in the array.

That is not true in the general case. At least I think it's not. Maybe
some of the real C gurus here can answer my original question about
whether it is undefined behaviour?

The two yield incompatible pointer types. I suppose a diagnostic is
required. I too await guru WRT undefined behavior.

--
Joe Wright mailto:jo****** **@comcast.net
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---

[infobahn]

Leon Brodskiy wrote:

Thanks for your answer.
I'm asking because I have found in a program the following call:

void f1(char inp[10])
{...}

void main()
A simple mistake. You meant:

int main(void)
{...
char str[10];
....
f1(&str);

This should be f1(str);

&str is a char (*)[10], not a char *, so it has the wrong type
for the f1() function.