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C: How come an array’s address is equal to its value?

I test in GCC 4.4.1 and I find &a=a. I can’t understand it. I think &a should be the address where stores the address of the array, it can’t be the same. Could someone give me a good explanation? Thanks very much.

Answer 1

An array is an object in memory. It has an address and a size. It’s also true that in certain contexts, an array decays into a pointer to its first element. So numerically, if both a and &a are compared as pointer values, they compare as equal, since they both point to the same address in memory. But they have different data types: a has type int[4] (“array 4 of int“), or int* (“pointer to int“) in certain contexts, whereas &a always has type int (*)[4] (“pointer to array 4 of int“).

 &a points here
 |
 V
+------+------+------+------+
| a[0] | a[1] | a[2] | a[3] |  sizeof(a) == 16
+------+------+------+------+
 ^
 |
 &a[0] also points here
 In certain contexts, 'a' means &a[0]

Hence, (void *)a == (void *)&a.

Also note that because a and &a point to different data types (and in particular, the pointed-to types have different sizes), doing pointer arithmetic will yield different results. a+1 will point to &a[1] (advances by one int value), whereas &a+1 will point to just past the end of the array at &a[4], since it advances by one “array 4 of int” unit:

                       &a+1 points here
                             |
                             V
+------+------+------+------+
| a[0] | a[1] | a[2] | a[3] |
+------+------+------+------+
        ^
        |
  a+1 points here