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] = '\0';
std::cout << pbuf << std::endl;
pbuf = new char[2]; //Lvalue Required ERROR
pbuf[0] = 'b';
pbuf[1] = '\0';
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] = '\0';
std::cout << pbuf << std::endl;
delete[] pbuf;
pbuf = new char[2];
pbuf[0] = 'b';
pbuf[1] = '\0';
std::cout << pbuf << std::endl;
delete[] pbuf;
return 0;
}
Why won't the first example work?
Thanks,
cpp 16 2121
"cppaddict" <he***@hello.com> wrote in message
news:k3********************************@4ax.com... I thought that:
char x[2];
made x into a pointer-to-char.
No. 'x' is an array.
An array is not a pointer.
A pointer is not an array. But how come the following code won't compile:
#include <iostream> /* for std::cout */
#include <ostream> /* for std::endl */
int main() { char pbuf[2]; pbuf[0] = 'a'; pbuf[1] = '\0'; std::cout << pbuf << std::endl;
pbuf = new char[2]; //Lvalue Required ERROR
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. :-)
pbuf[0] = 'b'; pbuf[1] = '\0'; std::cout << pbuf << std::endl; delete[] pbuf;
return 0; }
In contrast, the following code compiles and runs as expected:
Yes, because 'pbuf' is a pointer, to which you can assign
a value.
#include <iostream> /* for std::cout */
#include <ostream> /* for std::endl */
int main() { char *pbuf; pbuf = new char[2]; pbuf[0] = 'a'; pbuf[1] = '\0'; std::cout << pbuf << std::endl; delete[] pbuf;
pbuf = new char[2]; pbuf[0] = 'b'; pbuf[1] = '\0'; std::cout << pbuf << std::endl; delete[] pbuf;
return 0; }
Why won't the first example work?
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
cppaddict wrote: I thought that:
char x[2];
made x into a pointer-to-char. ...
No, 'x' here is of type 'char[2]'. The rest follows.
--
Best regards,
Andrey Tarasevich #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] = '\0';
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] = '\0';
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/de...em_getline.asp
Can you explain why the above two things work even though pointers and
arrays of different types?
Thanks,
cpp
"cppaddict" <he***@hello.com> wrote in message
news:o2********************************@4ax.com... #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] = '\0'; 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] = '\0'; 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/de...em_getline.asp
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.
Thanks, cpp
/ WP
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
cppaddict <he***@hello.com> wrote in message news:<k3********************************@4ax.com>. .. I thought that:
char x[2];
made x into a pointer-to-char.
No -- this defines x as an array of char.
But how come the following code won't compile:
int main() { char pbuf[2]; pbuf[0] = 'a'; pbuf[1] = '\0'; std::cout << pbuf << std::endl;
pbuf = new char[2]; //Lvalue Required ERROR
Because an array isn't a (modifiable) lvalue.
[ ... ]
Why won't the first example work?
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.
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
Le lundi 30 août 2004 à 06:03, cppaddict a écrit dans comp.lang.c++*: char pbuf[2]; pbuf[0] = 'a'; pbuf[1] = 'b'; pbuf[2] = '\0';
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
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
--
Karl Heinz Buchegger kb******@gascad.at
In article <k3********************************@4ax.com>,
cppaddict <he***@hello.com> wrote: I thought that:
char x[2];
made x into a pointer-to-char.
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.
On Mon, 30 Aug 2004 07:28:42 +0200
"William Payne" <mi**************@student.liu.se> wrote: 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.
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 Now would be a good time to become familiar with Chris Torek's, "The Rule". Search the Web for "Chris Torek The Rule". He explains in depth the difference between an array and a pointer.
Thanks for the reference.... very interesting.
cpp
Karl Heinz Buchegger wrote: 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
Also the address-of operator (&).
That means that &b is NOT char** but is a pointer to array 2 of char.
Brian Rodenborn
cppaddict <he***@hello.com> wrote: the following prints b:
int main() { char pbuf[2];
I guess you mean: char pbuf[3];
pbuf[0] = 'a'; pbuf[1] = 'b'; pbuf[2] = '\0'; std::cout << *(pbuf+1);
return 0; }
This makes it seem like a pointer.
When you use the name of an array in any expression other than
those listed below, it is as if you had written: &x[0].
The exceptions (I think..): sizeof x, &x, x++, ++x, --x, x--,
or where x is on the left-hand side of an assignment (=, +=, -=, etc.) This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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