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cppaddict
Guest
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 Posted: Mon Aug 30, 2004 2:47 am Post subject: What type is x in "char x[2];"? |
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I thought that:
char x[2];
made x into a pointer-to-char.
But how come the following code won't compile:
int main() {
char pbuf[2];
pbuf[0] = 'a';
pbuf[1] = ' ';
std::cout << pbuf << std::endl;
pbuf = new char[2]; //Lvalue Required ERROR
pbuf[0] = 'b';
pbuf[1] = ' ';
std::cout << pbuf << std::endl;
delete[] pbuf;
return 0;
}
In contrast, the following code compiles and runs as expected:
int main() {
char *pbuf;
pbuf = new char[2];
pbuf[0] = 'a';
pbuf[1] = ' ';
std::cout << pbuf << std::endl;
delete[] pbuf;
pbuf = new char[2];
pbuf[0] = 'b';
pbuf[1] = ' ';
std::cout << pbuf << std::endl;
delete[] pbuf;
return 0;
}
Why won't the first example work?
Thanks,
cpp
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Mike Wahler
Guest
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| Posted: Mon Aug 30, 2004 3:28 am Post subject: Re: What type is x in "char x[2];"? |
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"cppaddict" <hello (AT) hello (DOT) com> wrote
| Quote: | I thought that:
char x[2];
made x into a pointer-to-char.
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No. 'x' is an array.
An array is not a pointer.
A pointer is not an array.
| Quote: |
But how come the following code won't compile:
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#include <iostream> /* for std::cout */
#include <ostream> /* for std::endl */
| Quote: | int main() {
char pbuf[2];
pbuf[0] = 'a';
pbuf[1] = ' ';
std::cout << pbuf << std::endl;
pbuf = new char[2]; //Lvalue Required ERROR
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You Can't Do That. Arrays cannot be assigned to like that.
Only each individual element can be assigned.
Again, an array is not a pointer. A pointer is not an array.
I thought you'd been posting and reading here long enough
to know that by now. :-)
| Quote: | pbuf[0] = 'b';
pbuf[1] = ' ';
std::cout << pbuf << std::endl;
delete[] pbuf;
return 0;
}
In contrast, the following code compiles and runs as expected:
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Yes, because 'pbuf' is a pointer, to which you can assign
a value.
#include
#include <ostream> /* for std::endl */
| Quote: | int main() {
char *pbuf;
pbuf = new char[2];
pbuf[0] = 'a';
pbuf[1] = ' ';
std::cout << pbuf << std::endl;
delete[] pbuf;
pbuf = new char[2];
pbuf[0] = 'b';
pbuf[1] = ' ';
std::cout << pbuf << std::endl;
delete[] pbuf;
return 0;
}
Why won't the first example work?
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Because an array is not a pointer. Because arrays cannot
be assigned to. Because the return type of operator 'new'
is a pointer, so even if you could assign to an array,
it's the wrong type (it's not a pointer).
-Mike
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Andrey Tarasevich
Guest
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| Posted: Mon Aug 30, 2004 3:45 am Post subject: Re: What type is x in "char x[2];"? |
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cppaddict wrote:
| Quote: | I thought that:
char x[2];
made x into a pointer-to-char.
...
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No, 'x' here is of type 'char[2]'. The rest follows.
--
Best regards,
Andrey Tarasevich
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cppaddict
Guest
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| Posted: Mon Aug 30, 2004 4:03 am Post subject: Re: What type is x in "char x[2];"? |
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| Quote: | #include <iostream> /* for std::cout */
#include <ostream> /* for std::endl */
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Already had those. Just weren't in the post...
| Quote: | Again, an array is not a pointer. A pointer is not an array.
I thought you'd been posting and reading here long enough
to know that by now. 
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Indeed. I should have. I'm a little embarrassed, but my confusion
comes from some semi-legitimate sources. First, the fact that you can
do pointer arithmetic with arrays. For example, the following prints
b:
int main() {
char pbuf[2];
pbuf[0] = 'a';
pbuf[1] = 'b';
pbuf[2] = ' ';
std::cout << *(pbuf+1);
return 0;
}
This makes it seem like a pointer. Also, I've been working with the
Windows API, which often makes pointers and arrays seem
interchangeable. For example, consider this code for retrieving a
line from a Rich Edit control:
char pbuf[100];
pbuf[99] = ' ';
pbuf[0] = 99 //specifies max number of characters to retrieve;
SendMessage(hwndRichEdit,EM_GETLINE,0,(LPARAM)pbuf);
std::cout << "Line 1: " << pbuf << std::endl;
which works. The LPARAM is of type pointer I'm pretty sure. The docs
are here:
http://msdn.microsoft.com/library/default.asp?url=/library/en-us/shellcc/platform/commctls/editcontrols/editcontrolreference/editcontrolmessages/em_getline.asp
Can you explain why the above two things work even though pointers and
arrays of different types?
Thanks,
cpp
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William Payne
Guest
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| Posted: Mon Aug 30, 2004 5:28 am Post subject: Re: What type is x in "char x[2];"? |
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"cppaddict" <hello (AT) hello (DOT) com> wrote
| Quote: |
#include <iostream> /* for std::cout */
#include <ostream> /* for std::endl */
Already had those. Just weren't in the post...
Again, an array is not a pointer. A pointer is not an array.
I thought you'd been posting and reading here long enough
to know that by now. :-)
Indeed. I should have. I'm a little embarrassed, but my confusion
comes from some semi-legitimate sources. First, the fact that you can
do pointer arithmetic with arrays. For example, the following prints
b:
int main() {
char pbuf[2];
pbuf[0] = 'a';
pbuf[1] = 'b';
pbuf[2] = ' ';
std::cout << *(pbuf+1);
return 0;
}
This makes it seem like a pointer. Also, I've been working with the
Windows API, which often makes pointers and arrays seem
interchangeable. For example, consider this code for retrieving a
line from a Rich Edit control:
char pbuf[100];
pbuf[99] = ' ';
pbuf[0] = 99 //specifies max number of characters to retrieve;
SendMessage(hwndRichEdit,EM_GETLINE,0,(LPARAM)pbuf);
std::cout << "Line 1: " << pbuf << std::endl;
which works. The LPARAM is of type pointer I'm pretty sure. The docs
are here:
http://msdn.microsoft.com/library/default.asp?url=/library/en-us/shellcc/platform/commctls/editcontrols/editcontrolreference/editcontrolmessages/em_getline.asp
Can you explain why the above two things work even though pointers and
arrays of different types?
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It works because when an array is passed to a function it decays to a
pointer to its first element.
/ WP
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Sohail
Guest
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| Posted: Mon Aug 30, 2004 6:54 am Post subject: Re: What type is x in "char x[2];"? |
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char x[2];
means x is a charachter array of size 2 .
x is kind of a constant pointer and hence you cannot dynamically
point x to some other memory location.
whereas char *px; means px is a pointer to charachter, since it is not
constant you can dynamically make it point to any new memory location
allocated with new char[]
Hope this helps.
Sohail
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Jerry Coffin
Guest
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| Posted: Mon Aug 30, 2004 7:01 am Post subject: Re: What type is x in "char x[2];"? |
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cppaddict <hello (AT) hello (DOT) com> wrote
| Quote: | I thought that:
char x[2];
made x into a pointer-to-char.
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No -- this defines x as an array of char.
| Quote: | But how come the following code won't compile:
int main() {
char pbuf[2];
pbuf[0] = 'a';
pbuf[1] = ' ';
std::cout << pbuf << std::endl;
pbuf = new char[2]; //Lvalue Required ERROR
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Because an array isn't a (modifiable) lvalue.
[ ... ]
| Quote: | Why won't the first example work?
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Your code is ill-formed because your understanding about x was
incorrect.
As you've defined it above, x is an array. If you were to pass x as a
parameter to a function, then what would be passed would be a pointer
to the beginning of the array, but here you're not passing it as a
parameter. As it stands right now, you're trying to assign to x which
is an array, and assigning to an array simply isn't allowed.
Note that as a parameter, what you get is a pointer even if you use
array-style notation in the function declaration/definition. For
example:
void f(char x[2]) {
// This is well-formed code. The '2' above is entirely ignored
// so we can assign a pointer to a larger array, such as:
x = new char[20];
}
You can argue (and I'd agree) that this is usually a bad idea, but the
compiler won't do a thing to stop it. If you're feeling particularly
ornery, you can even do something like:
void f(char x[2]);
void f(char *x) { }
and get away with it -- at least with the compiler. Of course, if any
of your co-workers find out, it's your problem, not mine! :-)
--
Later,
Jerry.
The universe is a figment of its own imagination.
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David Michell
Guest
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| Posted: Mon Aug 30, 2004 7:48 am Post subject: Re: What type is x in "char x[2];"? |
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char pbuf[2];
is an array. An array is NOT same as a pointer.
It means the address the array points to is a constant, it cannot be changed.
Hence
pbuf = new char[2];
is WRONG
In this
char *pbuf;
pbuf is NOT an array, it is a pointer so its address can be changed.
Hence
pbuf = new char[2];
works fine.
hth
david michell
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Serge Paccalin
Guest
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| Posted: Mon Aug 30, 2004 9:01 am Post subject: Re: What type is x in "char x[2];"? |
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Le lundi 30 août 2004 à 06:03, cppaddict a écrit dans comp.lang.c++ :
| Quote: | char pbuf[2];
pbuf[0] = 'a';
pbuf[1] = 'b';
pbuf[2] = ' ';
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Now, wait a minute! You declare an array of two, then assign three
elements...
--
___________ 2004-08-30 11:00:46
_/ _ _`_`_`_) Serge PACCALIN -- sp ad mailclub.net
_L_) Il faut donc que les hommes commencent
-'(__) par n'être pas fanatiques pour mériter
_/___(_) la tolérance. -- Voltaire, 1763
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Karl Heinz Buchegger
Guest
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| Posted: Mon Aug 30, 2004 9:59 am Post subject: Re: What type is x in "char x[2];"? |
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cppaddict wrote:
| Quote: |
Can you explain why the above two things work even though pointers and
arrays of different types?
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You need to distinguish between what things *are* and how things are
*used*
If the name of an array is used without an indexing operation (*), then the
name of the array decays into a pointer to its first element. But note:
The name is still not a pointer, it is just used as if it were one.
Heck. Even array indexing is defined in terms of pointer operations:
char b[10];
b[5] is identical to *(b+5) by definition
(*) with sizeof beeing one exception that comes to my mind immediatly
--
Karl Heinz Buchegger
[email]kbuchegg (AT) gascad (DOT) at[/email]
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Daniel T.
Guest
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| Posted: Mon Aug 30, 2004 11:16 am Post subject: Re: What type is x in "char x[2];"? |
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In article <k355j0hoksvc0u5m9kthp6j48pvs637v1a (AT) 4ax (DOT) com>,
cppaddict <hello (AT) hello (DOT) com> wrote:
| Quote: | I thought that:
char x[2];
made x into a pointer-to-char.
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It's more like a const-pointer-to-char. Note, I'm not saying it *is*
such a beast, but its very much *like* one.
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Moritz Beller
Guest
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| Posted: Mon Aug 30, 2004 1:21 pm Post subject: Re: What type is x in "char x[2];"? |
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On Mon, 30 Aug 2004 07:28:42 +0200
"William Payne" <mikas493_no_spam (AT) student (DOT) liu.se> wrote:
| Quote: | Can you explain why the above two things work even though pointers
and arrays of different types?
It works because when an array is passed to a function it decays to a
pointer to its first element.
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Which also makes sizeof useless within the function. If you were to know
the array's elements, you'd need to add a parameter index:
foo(array,sizeof array/sizeof array[0]);
best regards
Moritz Beller
--
web http://www.4momo.de
mail momo dot beller at t-online dot de
gpgkey http://gpg.notlong.com
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Thomas Matthews
Guest
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Mike Wahler
Guest
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Default User
Guest
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| Posted: Mon Aug 30, 2004 5:14 pm Post subject: Re: What type is x in "char x[2];"? |
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Karl Heinz Buchegger wrote:
| Quote: |
cppaddict wrote:
Can you explain why the above two things work even though pointers and
arrays of different types?
You need to distinguish between what things *are* and how things are
*used*
If the name of an array is used without an indexing operation (*), then the
name of the array decays into a pointer to its first element. But note:
The name is still not a pointer, it is just used as if it were one.
Heck. Even array indexing is defined in terms of pointer operations:
char b[10];
b[5] is identical to *(b+5) by definition
(*) with sizeof beeing one exception that comes to my mind immediatly
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Also the address-of operator (&).
That means that &b is NOT char** but is a pointer to array 2 of char.
Brian Rodenborn
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