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(char *) to (const char *) is also dangerous but allowed?

 
 
lovecreatesbea...@gmail.com
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      04-09-2007
C stops the conversion from (char **) to (const char **). c-faq.com
sec 11.10 has explanation on this point. But, for example, even the
conversion from (char *) to (const char *) brings the same dangerous
as in the previous conversion. Why the latter simple but dangerous one
is allowed in C?

$ cat f1.c
int main(void)
{
const char c = 'a';
char *p;
const char *cp = p;

cp = &c;
*p = 'x'; /*line 8*/

return 0;
}
$ cc -Aa -g f1.c
$ ./a.out
Bus error(coredump)
$ gdb -q ./a.out core
Core was generated by `a.out'.
Program terminated with signal 10, Bus error.

warning: The shared libraries were not privately mapped; setting a
breakpoint in a shared library will not work until you rerun the
program.

#0 0x29d4 in main () at f1.c:8
8 *p = 'x';
(gdb) quit
$

 
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christian.bau
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      04-09-2007
On Apr 9, 10:53 am, "(E-Mail Removed)"
<(E-Mail Removed)> wrote:
> $ cat f1.c
> int main(void)
> {
> const char c = 'a';
> char *p;
> const char *cp = p;
>
> cp = &c;
> *p = 'x'; /*line 8*/
>
> return 0;}


That crash has nothing to do with const char*. Remove all the lines
involving cp and c, and the program will crash just the same. This is
just a case of stupid programmer error.

 
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Jens Thoms Toerring
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      04-09-2007
(E-Mail Removed) <(E-Mail Removed)> wrote:
> C stops the conversion from (char **) to (const char **). c-faq.com
> sec 11.10 has explanation on this point. But, for example, even the
> conversion from (char *) to (const char *) brings the same dangerous
> as in the previous conversion. Why the latter simple but dangerous one
> is allowed in C?


> $ cat f1.c
> int main(void)
> {
> const char c = 'a';
> char *p;
> const char *cp = p;


> cp = &c;
> *p = 'x'; /*line 8*/


There isn't any issue with conversions between char and const char
at all here. The simple and only problem is that you use a random
address to store a value in. 'p' is never initialized, so there's
no memory assigned to it that you own. Instead 'p' will point to
some random position in memory. But you try to store a value at
that memory location anyway. This invokes undefined behaviour and
from now on anything can happen. You may get a segmentation fault,
a bus error, it may even appear to work or the (in)famous nasal
daemons could make their appearance.

Regards, Jens
--
\ Jens Thoms Toerring ___ http://www.velocityreviews.com/forums/(E-Mail Removed)
\__________________________ http://toerring.de
 
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Malcolm McLean
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      04-09-2007

"(E-Mail Removed)" <(E-Mail Removed)> wrote in message
>C stops the conversion from (char **) to (const char **). c-faq.com
> sec 11.10 has explanation on this point. But, for example, even the
> conversion from (char *) to (const char *) brings the same dangerous
> as in the previous conversion. Why the latter simple but dangerous one
>?

It isn't dangerous to cast a pointer to a const *, as long as the data is
set up correctly. Going in reverse from const * to a plain pointer is
potentially dangerous, and is only allowed because of weaknesses in the
language arising from the fact that const was an afterthought.
--
Free games and programming goodies.
http://www.personal.leeds.ac.uk/~bgy1mm

 
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lovecreatesbea...@gmail.com
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      04-09-2007
On Apr 9, 4:03 am, (E-Mail Removed) (Jens Thoms Toerring) wrote:
> (E-Mail Removed) <(E-Mail Removed)> wrote:
> > C stops the conversion from (char **) to (const char **). c-faq.com
> > sec 11.10 has explanation on this point. But, for example, even the
> > conversion from (char *) to (const char *) brings the same dangerous
> > as in the previous conversion. Why the latter simple but dangerous one
> > is allowed in C?
> > $ cat f1.c
> > int main(void)
> > {
> > const char c = 'a';
> > char *p;
> > const char *cp = p;
> > cp = &c;
> > *p = 'x'; /*line 8*/

>
> There isn't any issue with conversions between char and const char
> at all here. The simple and only problem is that you use a random
> address to store a value in. 'p' is never initialized, so there's
> no memory assigned to it that you own. Instead 'p' will point to
> some random position in memory. But you try to store a value at
> that memory location anyway. This invokes undefined behaviour and
> from now on anything can happen. You may get a segmentation fault,
> a bus error, it may even appear to work or the (in)famous nasal
> daemons could make their appearance.


I'm sorry to make a wrong example. The pointer p wasn't initiated. p
and cp ever pointed to a same address, cp was re-assigned, p wasn't
changed and had a random address all the time. I understand it now.
Thank you for your time.

 
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Martin Ambuhl
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      04-09-2007
(E-Mail Removed) wrote:
> Why the latter simple but dangerous one
> is allowed in C?


Because the programmer is responsible for making sure that pointers have
legal values. Your program self-destructs for a reason completely
unrelated to the const-ness of the chars and and pointers-to-char.
>
> $ cat f1.c
> int main(void)
> {
> const char c = 'a';
> char *p;


p is not initialized.

> const char *cp = p;


p is still not initialized, and the value of cp is indeterminate,


> cp = &c;
> *p = 'x'; /*line 8*/

^^
dereferencing p, which has never been given a value.
>
> return 0;
> }

 
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Vallabha
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      04-11-2007
On Apr 9, 5:53 am, "(E-Mail Removed)"
<(E-Mail Removed)> wrote:
> C stops the conversion from (char **) to (const char **). c-faq.com
> sec 11.10 has explanation on this point. But, for example, even the
> conversion from (char *) to (const char *) brings the same dangerous
> as in the previous conversion. Why the latter simple but dangerous one
> is allowed in C?
>
> $ cat f1.c
> int main(void)
> {
> const char c = 'a';
> char *p;
> const char *cp = p;
>
> cp = &c;
> *p = 'x'; /*line 8*/
>
> return 0;}
>
> $ cc -Aa -g f1.c
> $ ./a.out
> Bus error(coredump)
> $ gdb -q ./a.out core
> Core was generated by `a.out'.
> Program terminated with signal 10, Bus error.
>
> warning: The shared libraries were not privately mapped; setting a
> breakpoint in a shared library will not work until you rerun the
> program.
>
> #0 0x29d4 in main () at f1.c:8
> 8 *p = 'x';
> (gdb) quit
> $


It's due to programming error.

However, you can change the value of const char using a pointer.

Ex:

#include <stdio.h>


int main ()
{
const char c = 'c';
char *p = &c;
*p = 'x';
printf("%c \n", c);
return 0;
}

However compiler thorws the warning of using a char pointer to a const
character.

$>cc del.c
"del.c", line 7: warning: assignment type mismatch:
pointer to char "=" pointer to const char

Cheers
-Vallabha


 
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Flash Gordon
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      04-11-2007
Vallabha wrote, On 11/04/07 06:30:

<snip>

> However, you can change the value of const char using a pointer.


You can attempt to, but it invokes undefined behaviour which means that
anything can happen, including it causing your in-laws to move in
permanently. The most likely results are either the program crashing or
working as you expect. Don't do it.

> Ex:
>
> #include <stdio.h>
>
>
> int main ()


Better to be explicit
int main(void)

> {
> const char c = 'c';
> char *p = &c;
> *p = 'x';
> printf("%c \n", c);
> return 0;
> }
>
> However compiler thorws the warning of using a char pointer to a const
> character.


Some compilers do, but this is not required and not all compiler will.
--
Flash Gordon
 
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Richard Heathfield
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      04-11-2007
Flash Gordon said:

> [...] it invokes undefined behaviour which means
> that anything can happen, including it causing your in-laws to move in
> permanently.


The UB war just got a little bit nastier.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at the above domain, - www.
 
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Charlton Wilbur
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      04-11-2007
>>>>> "RH" == Richard Heathfield <(E-Mail Removed)> writes:

RH> Flash Gordon said:
>> [...] it invokes undefined behaviour which means that anything
>> can happen, including it causing your in-laws to move in
>> permanently.


RH> The UB war just got a little bit nastier.

There's a reason the DS9000 was a commercial failure.

Charlton


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Charlton Wilbur
(E-Mail Removed)
 
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