Ian Lance Taylor wrote:
I am a little confused. The rule says:Joe Steeve <joe_steeve@xxxxxxx> 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."
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.
According to the rule, you cannot use the variable, declared as a const-qualified type, on the left hand side (lvalue, or assigned-to variable), but it does not say anything about the memory being modified.
Please correct me if I am wrong. Also, when you qualify the type of x to be a volatile, then the behaviour is as expected. So, there must be some other reason why the variable "int *ptr" is being discarded by Optimization. Check out that without optimization, you are able to change the contents. Also, I've tried this with a few other compilers (without optimization) and they all change the contents.
Harsha
Ian
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