CODEWITHSUNDEEP
c++pointerscasting
Corelli
Corelli

Reputation: 21

int* value of a void* address

So, I was messing with pointers, and I've got confused with some things. First, let me start with a basic example which beahaves like expected:

void* h = new char[16] {}; //pointer to a array of char, 16 bytes, and initilizes with 0
int* o = (int*)h; //gets the adress of the first byte
*o = 16; //sets the value
std::cout << *(int*)h << std::endl; //Prints the memory value
std::cout << *o; //Prints the memory value

and It prints this:

16
16

But this one doesn't output what I think it would:

    int* o = (int*)h+1; //gets the adress of the second byte
    *o = 16; //sets the value
    std::cout << *(int*)h+1 << std::endl; //Prints the memory value
    std::cout << *o; //Prints the memory value

But it outputs:

1
16

Shouldn't the two numbers be 16? As far as I know, by adding value to the pointer it increments the memory in bytes. So, is there something that I'm missing here?

Upvotes: 1

Views: 208

Answers (1)

rodrigo
rodrigo

Reputation: 98338

You have an issue with precedence of operators. Of all the operators you are using the highest precedence is that of the cast to (int*). So when you do (int*)h+1 you are actually doing ((int*)h)+1, and that is not the pointer to the second byte, but a pointer to the second integer, that is you are advancing sizeof(int) bytes.

Similarly with *(int*)h+1 you are actually doing (*(int*)h)+1, that is you are reading the first integer (that would be 0) and adding 1 to that integer (0 + 1 = 1). In this case you are not doing pointer arithmetic.

If you want to do proper pointer arithmentic you need a few parenthesis, but note that you cannot portably do pointer arithmentic with void *: use instead char *

Upvotes: 2

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