int p = *（int *）i和int p = *（int *）＆i之间的差异

Question

The following question was asked in a recent microsoft interview.

What is the difference between the two declarations?

int p=*(int*)i; 
int p=*(int*)&i; 

I think in the first one i is a pointer and in the second one i is a variable.
Is there anything else?

Answer 1

The first is taking the value contained in i , treating it as a pointer, and retrieving whatever int value is at that address (if possible).

The second takes the address of i , casts it to pointer to int, and retrieves the value at that address. If i is an int , it's equivalent to p=i; . If it's not, it's going to take the first CHAR_BIT *sizeof(int) bits starting at the address of i , and (attempt to) treat them as an int , and assign whatever value that creates to p .

Edit: and yes, as @R. Martinho Fernandes pointed out, if i has an overloaded operator & , it may do something rather different from any of the above (ie, instead of the address of i it'll start with whatever its operator & returns).

Answer 2

If you know the language then this question boils down to what is the difference betweeen

i

and

&i

And the answer is that in the first case it's i and in the second case it's address of i , and then you have all those conversions to act on either of these two.