Why is arr and &arr the same?

Question

I have been programming c/c++ for many years, but todays accidental discovery made me somewhat curious... Why does both outputs produce the same result in the code below? (<

Answer 1

They are not the same.

A bit more strict explanation:

arr is an lvalue of type int [3]. An attempt to use arr in some expressions like cout << arr will result in lvalue-to-rvalue conversion which, as there are no rvalues of array type, will convert it to an rvalue of type int * and with the value equal to &arr[0]. This is what you can display.

&arr is an rvalue of type int (*)[3], pointing to the array object itself. No magic here :-) This pointer points to the same address as &arr[0] because the array object and its first member start in the exact same place in the memory. That's why you have the same result when printing them.

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An easy way to confirm that they are different is comparing *(arr) and *(&arr): the first is an lvalue of type int and the second is an lvalue of type int[3].