An interesting problem about C++

zyzyis · 10-10-2004, 11:43 AM

Code:

#include <iostream>
using namespace std;

int main( int argc, char* argv[] )
{
    const int a=5;
    const int* aptr = &a;
    int* ptr = ( int* )(aptr);
    cout << ptr << " " << &a << " " << aptr << endl;
    *ptr = 3;
    cout<< *ptr << " " << a << " " << *aptr <<endl;
    return 0;
}

compile it with g++ and the result is
0xbffffb34 0xbffffb34 0xbffffb34
3 5 3

I know it's fool to write code like that in a really software, but

I'm really interested at how the g++ work with the const pointer typecast

anyone could tell me ?

thanks very much~~~

MadCactus · 10-10-2004, 12:11 PM

Why are you converting a_ptr when its already a pointer?

Mara · 10-10-2004, 12:43 PM

Haven't looked in the g++ source for this, but I think that 'const' just adds work to the compiler. It produces an error when the value is changed. But with special casting (pointer to pointer is enough) you can modify consts, so it's not very sophisticated.

aluser · 10-10-2004, 09:00 PM

I compiled the following with gcc

Code:

#include <stdio.h>

int main()
{
        const int a = 5;
        const int *aptr = &a;
        int *ptr = (int*)aptr;
        printf("%p %p %p\n", ptr, &a, aptr);
        *ptr = 3;
        printf("%d %d %d\n", *ptr, a, *aptr);
        return 0;
}

and found that, without any flags, it prints 3 3 3, but with -O1 through -O3 it prints 3 5 3.

This simple answer to the question is of course that writing to a const variable gives undefined results

You could get a better answer by using the -S switch to generate assembly output.. but I'm not good at reading that.

Mara · 10-11-2004, 02:38 PM

I checked with -O3 and it optimizes the second printf in a funny way. No pointers, but the values are passed to printf. The compiler assumes that aptr content doesn't change (it shouldn, right? it's a const int*) and uses its orginal value instead of getting what's currently kept in a.