Plain array
decays to a pointer to its first element, it's equal to &array[0]
. The first element also happens to start at the same address as the array itself. Hence &array == &array[0]
.
But it's important to note that the types are different:
- The type of
&array[0]
is (in your example) int*
.
- The type of
&array
is int(*)[5]
.
The relationship between &array[0]
and &array
might be easier if I show it a little more "graphically" (with pointers added):
+----------+----------+----------+----------+----------+
| array[0] | array[1] | array[2] | array[3] | array[4] |
+----------+----------+----------+----------+----------+
^
|
&array[0]
|
&array
As an extra addendum, array
decays to a pointer to its first element, that is array
decays to &array[0]
and will thus have the same type.
Things are different with pointers though. The pointer pArray
is pointing to some memory, the value of pArray
is the location of that memory. This is what you get when you use pArray
. It is also the same as &pArray[0]
.
When you use &pArray
you get a pointer to the pointer. That is, you get the location (address) of the variable pArray
itself. Its type is int**
.
Somewhat graphical with the pointer pArray
it would be something like this
+--------+ +-----------+-----------+-----------+-----------+-----------+-----+
| pArray | ----> | pArray[0] | pArray[1] | pArray[2] | pArray[3] | pArray[4] | ... |
+--------+ +-----------+-----------+-----------+-----------+-----------+-----+
^ ^
| |
&pArray &pArray[0]
[Note the ...
at the end of the "array", that's because pointers retains no information about the memory it points to. A pointer is only pointing to a specific location, the "first" element of the "array". Treating the memory as an "array" is up to the programmer.]
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