Why do these two C++ statements give two different locations in memory

Question

With the following code, two different addresses are printed out. Can anyone explain why? I'm not the most proficient with pointers but I presumed that they would give the same address.

#include <iostream>

int main() {

    float (*arrayName)[2][4][4];

    auto address1 = (**arrayName)[1]; // 0x10
    auto address2 = (*arrayName)[1][0]; // 0x40
    std::cout << "Address1: " << address1 << std::endl;
    std::cout << "Address2: " << address2 << std::endl;
}

0x40 is the address I'm looking for. Just don't understand why I get two different answers.

I was expecting both to give me an answer of 0x40 however one doesn't. Tried this code on an online C++ compiler and both gave me the address of 0x10. However on my development RHEL7 vm, I get 0x10 and the expected 0x40 address.

Answer 1

As for what's really going on, please learn that for any pointer or array p and index i, the expression p[i] is exactly equal to *(p + i).

From that it follows that p[0] must be equal to *(p + 0) which of course is equal to *(p) which is the same as *p.

And from that we can learn that *arrayName is the same as arrayName[0], from which follows that **arrayName is arrayName[0][0].

With this information it's easy to see that (*arrayName)[1][0] is different from (**arrayName)[1]:

• (*arrayName)[1][0] is the same as arrayName[0][1][0].

• (**arrayName)[1] is the same as arrayName[0][0][1].

But remember, because arrayName isn't assigned or initialized with a specific pointer, it can't be dereferenced.