Difference between array and &array[0]

Question

int array[] = {1,2,3,4};

As I understand, array is just a pointer to &array[0]

But how come then sizeof(array); knows size of an array and not just that it's only 4 byte number?

Answer 1

Although the name of the array does become a pointer to its initial member in certain contexts (such as passing the array to a function) the name of the array refers to that array as a whole, not only to the pointer to the initial member.

This manifests itself in taking the size of the array through sizeof operator.

Answer 2

Except when it is the operand of the sizeof or unary & operators, or is a string literal being used to initialize another array in a declaration, an expression of type "N-element array of T" will be converted ("decay") to an expression of type "pointer to T, and the value of the expression will be the address of the first element of the array.

Arrays in C don't store any metadata about their size or anything else, nor is any storage set aside for any sort of pointer. They're laid out pretty much as follows:

     +---+
arr: | 1 | arr[0]
     +---+
     | 2 | arr[1]
     +---+
     | 3 | arr[2]
     +---+
     | 4 | arr[3]
     +---+

There's no separate storage for a variable arr apart from the array elements themselves. As you can see, the address of arr and the address of arr[0] are the same. This is why the expressions arr, &arr, and &arr[0] all give you the same value (the address of the first element), even though the types of those expressions are different.

Except when the operand is a variable-length array, the result of sizeof is computed at compile time, and the compiler treats array operands as arrays in those circumstances; otherwise, it treats the array expression as a pointer to the first element.