Question

I got this weird behavior of my programm, that i cant figure out. My professor showed me a flaw in my programm, where i just copy a char pointer when i construct an object instead of making a new copy of the whole array, so you can fool around with it. He demonstrated this with similar code like this.

For the code:

char sweat[] ="Sweater";
warenkorb = new WareImKorb(new Textil (205366,4.2,sweat,40),2,warenkorb);
sweat[0] = '\0';

now if i instead make it:

char* sweat ="Sweater";

the program runs fine till i try sweat[0] = ‘\0’;
It simply crahes then.

However this works:
char cc[] =”Sweater”;
char* sweat = cc;

It is really bugging me, that i dont understand, why version 1 does not work.
Hope you guys can help me out, or else i will go crazy wondering about this.

Answer 1

char* sweat ="Sweater";
sweat[0] = '\0'

Here sweat points to a CONSTANT data. “Sweater” is const literal data, residing somewhere in read-only memory, and sweat points to this data as such. It doesn’t make a copy of it. So when you do sweat[0]='\0', it tries to change first character of the CONSTANT data. Hence the error. By the way, a good compiler should give warning if you don’t write const in your declaration, as const char* sweater = "Sweater". See the warning here : http://www.ideone.com/P47vv

But when you write char sweat[] = "Sweater", an array of char is created, copying the data from the CONSTANT data which is "Sweater"; that array’s element itself is modifiable!

---

Lets see an interesting thing: since in the first case, it doesn’t make a copy of the const data, so no matter how many variables you declare (all pointing to the same data), the address would be same for all variables. See this:

#include<cstdio>
int main() {
        char* sweat  ="Sweater";        //notice the warning
        const char* another ="Sweater"; //no warning!
        std::printf("%p\n", sweat);     //print the address
        std::printf("%p\n", another);   //print the address
        return 0;
}

Output:

0x8048610
0x8048610

Means, both printfs print the same address!

See yourself here : http://www.ideone.com/VcyM6