assigning to const int via pointer
Gabriel Dos Reis
Fri Dec 31 11:08:00 GMT 2004
Sriharsha <firstname.lastname@example.org> writes:
| Ian Lance Taylor wrote:
| >Joe Steeve <email@example.com> writes:
| >>The following code assigns a value to a `const int` via a
| >>#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.
More information about the Gcc-help