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Julián Albo
Guest
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 Posted: Wed Jun 25, 2003 6:21 pm Post subject: Re: char * vs char[] |
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C Wood escribió:
| Quote: | void some_func() {
char temp[] = "prepend"; /*This one*/
char *temp= "prepend"; /*Or this one*/
strcat(temp," before this");
}
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None of them. char temp []= "prepend"; allocates just the space needed
for the characters of the string plus the zero terminator. If you strcat
to it you are out of space.
Regards.
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John Harrison
Guest
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| Posted: Wed Jun 25, 2003 6:25 pm Post subject: Re: char * vs char[] |
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"C Wood" <cdotrdotwood (AT) verizondotnet (DOT) removeme> wrote
| Quote: |
Dear group,
I forgot which declaration is constant. Intentions are:
void some_func() {
char temp[] = "prepend"; /*This one*/
char *temp= "prepend"; /*Or this one*/
strcat(temp," before this");
}
Thanks...
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The second, because all you've declared is a pointer which points at a
string literal, and string literals are read-only.
Despite appearances the first doesn't involve a string literal, its a
confusing short hand for this.
char temp[] = {'p', 'r', 'e', 'p', 'e', 'n', 'd', ' ' };
Obviously if you declare a non-const array, you can modify it.
Your strcat is illegal in both cases, nothing to do with const, instead you
are writing over memory you haven't allocated in any way.
john
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Thomas Matthews
Guest
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| Posted: Wed Jun 25, 2003 6:45 pm Post subject: Re: char * vs char[] |
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C Wood wrote:
| Quote: | Dear group,
I forgot which declaration is constant. Intentions are:
void some_func() {
char temp[] = "prepend"; /*This one*/
char *temp= "prepend"; /*Or this one*/
strcat(temp," before this");
}
Thanks...
|
In either case the literal text "prepend" is constant.
Also, you can't have a variable declared with two
different types.
The first declaration declares an array of char.
The array is initialized with the characters from
the text literal. The size of the array is determined
by the length of the literal.
In the second declaration, you are declaring a pointer
to char which initialized to point to the literal.
The pointer can be modified. If the target of the
pointer is modified, undefined behavor is invoked.
The second declaration causes the literal to have
a physical address. In the first declaration, the
literal doesn't have to have a physical address;
the literal can disappear after the declaration
is executed.
--
Thomas Matthews
C++ newsgroup welcome message:
http://www.slack.net/~shiva/welcome.txt
C++ Faq: http://www.parashift.com/c++-faq-lite
C Faq: http://www.eskimo.com/~scs/c-faq/top.html
alt.comp.lang.learn.c-c++ faq:
http://www.raos.demon.uk/acllc-c++/faq.html
Other sites:
http://www.josuttis.com -- C++ STL Library book
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C Wood
Guest
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| Posted: Wed Jun 25, 2003 6:53 pm Post subject: Re: char * vs char[] |
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"Thomas Matthews" <thomas_matthews (AT) sbcglobal (DOT) net> wrote
: C Wood wrote:
: In either case the literal text "prepend" is constant.
: Also, you can't have a variable declared with two
: different types.
:
: The first declaration declares an array of char.
: The array is initialized with the characters from
: the text literal. The size of the array is determined
: by the length of the literal.
:
: In the second declaration, you are declaring a pointer
: to char which initialized to point to the literal.
: The pointer can be modified. If the target of the
: pointer is modified, undefined behavor is invoked.
:
: The second declaration causes the literal to have
: a physical address. In the first declaration, the
: literal doesn't have to have a physical address;
: the literal can disappear after the declaration
: is executed.
:
I know, I tossed a fast sample using syntax I don't normally use and
totally wrote trash uncompilable code.
What I really wanted was a better way than this:
void func() {
char temp[8];
temp[0] = 'P';
temp[1] = 0;
strcat(temp,"asdf");
}
Application is tossing in drive letters and such before filenames, in a
clean and readable manner. Why worry about what works? It just annoying to
do it the slow way, when it seems it could be preinitiailized. Probably
worried about something that makes no difference anyways. Premature
opt.............l.
That way the temp would be:
"Pasdf"
Looks like the method:
char temp[8] = {'P',0,0,0,0,0,0,0};
is the best, or the way I've been doing it is fine also.
Back to work it is then...........
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Corey Murtagh
Guest
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| Posted: Wed Jun 25, 2003 7:15 pm Post subject: Re: char * vs char[] |
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C Wood wrote:
<snip>
| Quote: |
Looks like the method:
char temp[8] = {'P',0,0,0,0,0,0,0};
is the best, or the way I've been doing it is fine also.
Back to work it is then...........
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Err...
char temp[8] = "P";
or
char temp[8] = {'P', 0};
Works just as well, except I can't guarantee that the rest of temp past
the first 2 bytes will be initialized to anything useful (like nuls for
example). Saves on all that typing of nuls at least.
--
Corey Murtagh
The Electric Monk
"Quidquid latine dictum sit, altum viditur!"
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Ron Natalie
Guest
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| Posted: Wed Jun 25, 2003 7:33 pm Post subject: Re: char * vs char[] |
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"Corey Murtagh" <emonk (AT) slingshot (DOT) co.nz.no.uce> wrote
| Quote: |
char temp[8] = {'P', 0};
Works just as well, except I can't guarantee that the rest of temp past
the first 2 bytes will be initialized to anything useful (like nuls for
example). Saves on all that typing of nuls at least.
|
Actually all you need is
char temp[8] = { 'P' };
The other seven spots are default initialized (zero'd).
If you specify fewer initializers than the size of the array, the rest are
default initialized.
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Thomas Matthews
Guest
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| Posted: Fri Jun 27, 2003 2:50 pm Post subject: Re: char * vs char[] |
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C Wood wrote:
| Quote: | "Gavin Deane" <deane_gavin (AT) hotmail (DOT) com> wrote in message
news:6d8002d0.0306251633.19d47d26 (AT) posting (DOT) google.com...
: "C Wood" <cdotrdotwood (AT) verizondotnet (DOT) removeme> wrote in message
news:XcmKa.12914$Xb.2040 (AT) nwrddc02 (DOT) gnilink.net...
: Any reason why std::string doesn't work for you?
: GJD
The reason is that compiler extensions call this function. Any
discussion of that is Off Topic here.
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Here is an example of using string and some functions/methods that
require char *:
#include <fstream>
#include <string>
#include <cstdlib>
using namespace std;
const string file_name = "my_data.txt";
int main(void)
{
ofstream my_file(file_name.c_str());
if (my_file)
{
my_file << "Hello." << endl;
cout << "File written." << endl;
}
else
{
cout << "Error opening file." << endl;
}
return EXIT_SUCCESS;
}
I would hope that in the next iteration of the standard,
they allow std::string as a parameter to standard methods
and functions that require char *.
--
Thomas Matthews
C++ newsgroup welcome message:
http://www.slack.net/~shiva/welcome.txt
C++ Faq: http://www.parashift.com/c++-faq-lite
C Faq: http://www.eskimo.com/~scs/c-faq/top.html
alt.comp.lang.learn.c-c++ faq:
http://www.raos.demon.uk/acllc-c++/faq.html
Other sites:
http://www.josuttis.com -- C++ STL Library book
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Rayiner Hashem
Guest
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| Posted: Fri Jun 27, 2003 8:16 pm Post subject: Re: char * vs char[] |
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"C Wood" <cdotrdotwood (AT) verizondotnet (DOT) removeme> wrote
| Quote: | Dear group,
I forgot which declaration is constant. Intentions are:
void some_func() {
char temp[] = "prepend"; /*This one*/
char *temp= "prepend"; /*Or this one*/
strcat(temp," before this");
}
Thanks...
If you have the STL available (and you should, you can use std::string |
in an OS kernel without any more runtime support than malloc/free  ,
you can do this:
char* some_func()
{
std::string str("prepend");
str += " before this";
return strdup(str.c_str()); // this will give the caller a copy of
the combined string
}
Even if you get one part or the other of the string as straight
C-strings, you can use the appropriate constructors for std::string to
handle the conversion under the hood.
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Ron Natalie
Guest
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| Posted: Fri Jun 27, 2003 9:06 pm Post subject: Re: char * vs char[] |
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"Rayiner Hashem" <heliosc (AT) mindspring (DOT) com> wrote
| Quote: | return strdup(str.c_str()); // this will give the caller a copy of
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strdup is not a standard function.
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