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

[lovecreatesbea...]

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
$

[David Thompson]

On Wed, 11 Apr 2007 07:48:49 +0100, Flash Gordon
<sp**@flash-gordon.me.ukwro te:

Vallabha wrote, On 11/04/07 06:30:

const char c = 'c';
char *p = &c;

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.

Well, removing const or volatile qualification from a pointer is a
constraint violation, and conforming implementation is required to
issue a diagnostic. Whether that diagnostic is a warning, and how well
it identifies the problem, is 'only' Quality of Implementation.

- formerly david.thompson1 || achar(64) || worldnet.att.ne t