How does de-referencing work for pointer to an array?

Question

Pointer to array of elements when dereferenced return an address. Since it is holding the address of the first element of the array, dereferencing it should return a value.

Answer 1

The question was

"Why does *ptr return the base address of the array, shouldn't it return the value at that address?"

It does return the value at that address, which is the array arr.

Think about a queue of people: you can point to the first person and say "that person over there" or you can point to the same person and say "that queue over there". There are 2 things at the same location: a person and a queue. Same thing happens with arrays: person * for that "person over there" and person (*)[42] "for that queue of 42 people". If you dereference a pointer to queue, you get a queue. If you take the first from the queue you get a person.

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But then the array itself will decay to address to the first element when it is given as an argument to printf. Thus here,

int arr[] = { 3, 5, 6, 7, 9 }; 
int (*ptr)[5] = &arr;     

// undefined behaviour really, all pointers should be cast to void *
printf("%p %p %p %p", *ptr, &ptr[0], arr, &arr[0]);

all these 4 expressions will result in pointer to int, and the value is the address of the first element in the array (the address of 3 in arr).