Pointer typecast char to int

[ram619]

Hello, in this code snippet I am not able to understand line 8 and 10. Here at line 8, char* is getting typecasted as int* then again, getting typecasted as char*. Please explain.

#include<stdio.h>

int main()
{
        int arr[3] = {2,3,4};
        char *p;
        p = arr;
        p = (char*)((int*) (p));
        printf("%d \n",*p);
        p = (int*) (p+1);
        printf("%d \n",*p);
}

[Ancient Dragon]

line 8 is nonsense -- no one in his/her right mind would do such a thing. Useless statement.

line 7 might be a problem without typecase.
p = (char*)arr;

line 9 will NOT print the value of arr[0] as you might expect. p is a char pointer, so *p will return only the first byte of an integer.

[sepp2k]

That line makes no sense, that's why you don't understand it. As you said, it's casting to int* and then back to char*. It could just as well be written p = p; (or just left out altogether because obviously that's a noop).

Line 7 should be p = (char*) arr;. You need a cast there because only void pointers can be implicitly converted to other pointer types - other pointers or arrays can't (though some (most?) compilers allow it with a warning).

The cast on line 10 doesn't make much sense either. Since p has type char*, this line relies on the int pointer being implicitly converted back to char*, which, as I said regarding line 7, is illegal. And if it was illegal it would be the same kind of pointless to-int*-and-back casting we see on line 8. I.e. it calculates p+1 (using the type char*) and then casts that to int* and then right back to char*. So line 10 is the same as p = p + 1; (except the latter does not rely on an illegal implicit cast).

PS: Your main function is missing a return statement.

[ram619]

Thanks, post solved :)

[Razahcy]

what purpose of typecasting in malloc function?