Why can i change the value of a constant (const char * ) trough a pointer?

Question

The title says it all. How am I able to change the value of a constant? Also is that the same as when you change the value of an element at index X of a constant array?

    #include<iostream>
    int main(){
        const char* y = "original";
        auto *p = &y;
        *p = "modified";
        std::cout<<y<<"\n";
        //outputs modified
        system("pause");
        return 0;
    }

hmmm i get that, but then at the end of the day does a constant string/char make sense anymore ? — Shish Kebab, Jul 29 '19 at 15:04
Yes it does but there is a difference btw constant pointer and pointer to const data. Also there is difference btw modifying pointer (make it point to different location) and modifying data it points to. — Slava, Jul 29 '19 at 15:08
Yes it makes sense. You can't change a literal but that's not what your code does, and `y` is not a constant... — Lightness Races in Orbit, Jul 29 '19 at 15:11
I that what you wanted? ```char * const y = "original";``` gives ```error: assignment of read-only location ‘* p’``` — Guillaume D, Jul 29 '19 at 15:13
ok i kind of got it with @songyuanyao's explanation now, thanks guys — Shish Kebab, Jul 29 '19 at 15:19

songyuanyao · Accepted Answer · 2019-07-29T15:21:17.837

Note that y is a non-const pointer (to const). You're not modifying the const part, i.e. the const char pointed by y, but y itself; this is valid. BTW modifying y through the pointer p is just same as

const char* y = "original";
y = "modified";  // this is well-formed

On the other hand,

*y = 'm';  // this is ill-formed

If you make y const, then you might get the diagnostic you expected.

const char* const y = "original";
y = "modified";  // ill-formed
*y = 'm';        // ill-formed

Why can i change the value of a constant (const char * ) trough a pointer?

1 Answers1