A problem about char pointer ?

**Victor Bazarov** · Mar 10 '08, 03:15 PM

Re: A problem about char pointer ?

comp.lang.c++ wrote:

this is a sample example about this question
#include<stdio. h>
void chg(char* t)
{
char *s=t;
char p=*t;
while(*t++=*++s );
*--t=p;
}
int main()
{
//char * t="abcde";
char t[]="abcde";
chg(t);
printf(t);
printf("\n");
return 0;
}
I want know why the function chg() cann't work when define t with
char*, and what's the different
between char* and char[]?

The difference is relatively obscure, unfortunately. When you declare
't' to be an array (char t[]), then each element of 't' is allowed to
be modified, IOW it's OK to modify it. When you declare 't' as a mere
pointer, and initialise it with a literal, then 't' points to the
non-modifiable memory, and any attempt to change the value of elements
of 't' has _undefined_beha viour_.

thanks?

Yes, please.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

**Jim Langston** · Mar 10 '08, 09:15 PM

Re: A problem about char pointer ?

Victor Bazarov wrote:

comp.lang.c++ wrote:

>this is a sample example about this question
>#include<stdio .h>
>void chg(char* t)
>{
> char *s=t;
> char p=*t;
> while(*t++=*++s );
> *--t=p;
>}
>int main()
>{
> //char * t="abcde";
> char t[]="abcde";
> chg(t);
> printf(t);
> printf("\n");
> return 0;
>}
>I want know why the function chg() cann't work when define t with
>char*, and what's the different
>between char* and char[]?

>
The difference is relatively obscure, unfortunately. When you declare
't' to be an array (char t[]), then each element of 't' is allowed to
be modified, IOW it's OK to modify it. When you declare 't' as a mere
pointer, and initialise it with a literal, then 't' points to the
non-modifiable memory, and any attempt to change the value of elements
of 't' has _undefined_beha viour_.
>

>thanks?

>
Yes, please.

Just to reiterate, the problem isn't, per se, that t is defined as a char*,
but where t is pointing to.

For example,

char x[] = "abcde";
char* t = x;
chg(t);

with your code would work. Because t is now not pointing to constant
memory.

--
Jim Langston
tazmaster@rocke tmail.com

**Paul Brettschneider** · Mar 10 '08, 11:05 PM

Re: A problem about char pointer ?

Victor Bazarov wrote:

comp.lang.c++ wrote:

>this is a sample example about this question
>#include<stdio .h>
>void chg(char* t)
>{
> char *s=t;
> char p=*t;
> while(*t++=*++s );
> *--t=p;
>}
>int main()
>{
> //char * t="abcde";
> char t[]="abcde";
> chg(t);
> printf(t);
> printf("\n");
> return 0;
>}
>I want know why the function chg() cann't work when define t with
>char*, and what's the different
>between char* and char[]?

>
The difference is relatively obscure, unfortunately. When you declare
't' to be an array (char t[]), then each element of 't' is allowed to
be modified, IOW it's OK to modify it. When you declare 't' as a mere
pointer, and initialise it with a literal, then 't' points to the
non-modifiable memory, and any attempt to change the value of elements
of 't' has _undefined_beha viour_.

Reminds me of my first paid coding job, which consisted of making a large
code base containing myriads of goodies like
char *s;
[...]
return strncpy(" ", s, 5);
and lots of global variables reentrant, because it was to be used in a
shared library.

I don't miss those times. ;)

**comp.lang.c++** · Mar 11 '08, 01:35 AM

Re: A problem about char pointer ?

On Mar 11, 5:06 am, "Jim Langston" <tazmas...@rock etmail.comwrote :

Victor Bazarov wrote:

comp.lang.c++ wrote:

this is a sample example about this question
#include<stdio. h>
void chg(char* t)
{
char *s=t;
char p=*t;
while(*t++=*++s );
*--t=p;
}
int main()
{
//char * t="abcde";
char t[]="abcde";
chg(t);
printf(t);
printf("\n");
return 0;
}
I want know why the function chg() cann't work when define t with
char*, and what's the different
between char* and char[]?

>

The difference is relatively obscure, unfortunately. When you declare
't' to be an array (char t[]), then each element of 't' is allowed to
be modified, IOW it's OK to modify it. When you declare 't' as a mere
pointer, and initialise it with a literal, then 't' points to the
non-modifiable memory, and any attempt to change the value of elements
of 't' has _undefined_beha viour_.

>

thanks?

>

Yes, please.

>
Just to reiterate, the problem isn't, per se, that t is defined as a char*,
but where t is pointing to.
>
For example,
>
char x[] = "abcde";
char* t = x;
chg(t);
>
with your code would work. Because t is now not pointing to constant
memory.
>
--
Jim Langston
tazmas...@rocke tmail.com

thanks ,I think I konw .Because "abcde" is a constant ,t just point to
it.and so t should be a constant pointer in fact.and char t[]="abcde"
is declare a array and give it's elements vuales by "abcde" , and t
is point to the first element's address.and I want to know how to
judge if the pointer is point to a constant in my function ? thanks!

**Triple-DES** · Mar 11 '08, 05:55 AM

Re: A problem about char pointer ?

On 11 Mar, 02:27, "comp.lang. c++" <cnhenu...@gmai l.comwrote:

thanks ,I think I konw .Because "abcde" is a constant ,t just point to
it.and so t should be a constant pointer in fact.and char t[]="abcde"
is declare a array and give it's elements vuales by "abcde" , and t
is point to the first element's address.and I want to know how to
judge if the pointer is point to a constant in my function ? thanks!

Since your function takes a pointer to char, it should not be pointing
to a constant. Just make sure to use const char pointers when dealing
with literals. Or preferably, std::string

DP

**Jim Langston** · Mar 14 '08, 10:15 AM

Re: A problem about char pointer ?

comp.lang.c++ wrote:

On Mar 11, 5:06 am, "Jim Langston" <tazmas...@rock etmail.comwrote :

>Victor Bazarov wrote:

>>comp.lang.c ++ wrote:
>>>this is a sample example about this question
>>>#include<std io.h>
>>>void chg(char* t)
>>>{
>>> char *s=t;
>>> char p=*t;
>>> while(*t++=*++s );
>>> *--t=p;
>>>}
>>>int main()
>>>{
>>> //char * t="abcde";
>>> char t[]="abcde";
>>> chg(t);
>>> printf(t);
>>> printf("\n");
>>> return 0;
>>>}
>>>I want know why the function chg() cann't work when define t with
>>>char*, and what's the different
>>>between char* and char[]?

>>

>>The difference is relatively obscure, unfortunately. When you
>>declare 't' to be an array (char t[]), then each element of 't' is
>>allowed to be modified, IOW it's OK to modify it. When you declare
>>'t' as a mere pointer, and initialise it with a literal, then 't'
>>points to the non-modifiable memory, and any attempt to change the
>>value of elements of 't' has _undefined_beha viour_.

>>

>>>thanks?

>>

>>Yes, please.

>>
>Just to reiterate, the problem isn't, per se, that t is defined as a
>char*, but where t is pointing to.
>>
>For example,
>>
>char x[] = "abcde";
>char* t = x;
>chg(t);
>>
>with your code would work. Because t is now not pointing to constant
>memory.
>>
>--
>Jim Langston
>tazmas...@rock etmail.com

thanks ,I think I konw .Because "abcde" is a constant ,t just point to
it.and so t should be a constant pointer in fact.and char t[]="abcde"
is declare a array and give it's elements vuales by "abcde" , and t
is point to the first element's address.and I want to know how to
judge if the pointer is point to a constant in my function ? thanks!

Use constant correctness in your code. If anything is constant, declare it
as such. The following will not compile with your code:

int main()
{
const char* t="abcde";
chg(t);
}

The error that MSVC++ .net 2003 gives me is:
error C2664: 'chg' : cannot convert parameter 1 from 'const char *' to 'char
*'

The reason you are allowed to have a char pointer point to non constant
memory in the first place is because it was allowed in C. For backwards
compatability. As such the programmer has to be a bit more careful with
what they are doing.

Use the const keyword wherever you can.
--
Jim Langston
tazmaster@rocke tmail.com

A problem about char pointer ?

A problem about char pointer ?

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