Re: assigning to const int via pointer

[Gabriel Dos Reis <gdr at integrable-solutions dot net>]

Sriharsha <sriharsha.v@redpinesignals.com> writes:

| Ian Lance Taylor wrote:
| 
| >Joe Steeve <joe_steeve@gmx.net> writes:
| >
| >
| >>The following code assigns a value to a `const int` via a
| >>pointer.,
| >>
| >>#include <stdio.h>
| >>
| >> int main()
| >>{
| >>  const int x=5;
| >>  int *ptr;
| >>  ptr = &x;
| >>  *ptr = 10;
| >>  printf("%d",x);
| >>}
| >>
| >>The code gives `10` for the following compilation
| >>
| >>    $gcc -o test test.c
| >>
| >>It gives `5` when using optimisations switches.,
| >>
| >>    $gcc -o test -O2 test.c
| >>
| >> Feature or bug or any explanation for this?
| >
| >When you declare that the variable is const, you are declaring that
| >the value does not change.  When you do change it, you are using
| >undefined behaviour.  When the compiler sees undefined behaviour, it
| >does not behave predictably.
| >
| >ISO C99 6.7.3: "If an attempt is made to modify an object defined with
| >a const-qualified type through use of an lvalue with
| >non-const-qualified type, the behavior is undefined."
| >
| I am a little confused. The rule says:
| If an attempt is made to modify an object, defined with a
| const-qualified type, through use of an lvalue with a
| non-const-qualified type, the behavior is undefined.
| 
| Now, in the above, program, we are not trying to alter the value of
| the variable x as follows:
| x = 10;
| 
| But we are trying to alter the contents of a memory location, which
| happens to be where the variable 'x' refers to,  by using a pointer,
| which is defined behaviour.

which is precisely what the rule outlaws.  

-- Gaby