Why does dereferencing an array or not result in the same address?

Question

In C++, I wrote the following simple main:

int main() {
    char test[100];
    void* a = (void*) test;
    void* b = (void*) &test;

    std::cout << a << " " << b << std::endl;

    return 0;
}

And it gives me the same result for a and b. Why is this? I would expect from the notation that the second be the address of the first..

Because in the `a` line, `test` is shorthand for `&(test[0])`, and what you are printing out is the fact that the first element of an array has the same address as the whole array. — M.M, Feb 17 '15 at 21:54
possible duplicate of [How come an array's address is equal to its value in C?](http://stackoverflow.com/questions/2528318/how-come-an-arrays-address-is-equal-to-its-value-in-c) — R Sahu, Feb 17 '15 at 21:58
I'm not really a fan of closing C++ questions as duplicates of C questions (even if the principle is the same in both languages) — M.M, Feb 17 '15 at 22:36

score 6 · Answer 1 · edited May 23 '17 at 12:28

In C++, arrays are converted to pointer to first element of the array. test is pointer to first element test[0]. &test is the address of entire array test. Although, the type of test and &test are different, their values are same and that's why you are getting the same value.

For example

int a[3] = {5, 4, 6};

Look at the diagram below:

enter image description here

For detailed explanation read this answer.

Why does dereferencing an array or not result in the same address?

1 Answers1