A problem about char pointer ?

[comp.lang.c++]

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[]?
thanks?

[Victor Bazarov]

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_behaviour_.

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]

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_behaviour_.

>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
ta*******@rocketmail.com

[Paul Brettschneider]

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_behaviour_.

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++]

On Mar 11, 5:06 am, "Jim Langston" <[e-mail]:

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_behaviour_.

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
[e-mail]

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]

On 11 Mar, 02:27, "comp.lang.c++" <[e-mail]:

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]

comp.lang.c++ wrote:

On Mar 11, 5:06 am, "Jim Langston" <[e-mail]:

>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_behaviour_.

>>>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
[e-mail]

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
ta*******@rocketmail.com