Address of array

问题:

int t[10];

int * u = t;

cout << t << " " << &t << endl;

cout << u << " " << &u << endl;

output:

0045FB88 0045FB88
0045FB88 0045FB7C

The output for u makes sense.

I understand that t and &t[0] should have the same value, but how come &t is also the same? What does &t actually mean?

回答1:

When t is used on its own in the expression, an array-to-pointer conversion takes place, this produces a pointer to the first element of the array.

When t is used as the argument of the & operator, no such conversion takes place. The & then explicitly takes the address of t (the array). &t is a pointer to the array as a whole.

The first element of the array is at the same position in memory as the start of the whole array, and so these two pointers have the same value.

回答2:

Actual type of t is int[10], so &t is the address of array.
Also, int[] implicitly converts to int*, so t converts to address of array the first element of array.

回答3:

There is no variable called t, since you can't change it. The name t simply refers to the address of the first element (and also has a size associated with it). Thus, taking the address of the address doesn't really make sense, and C "collapses" it into just being the address.

The same sort of thing happens for the case of functions:

int foo(void)
{
  return 12;
}

printf("%p and %p\n", (void *) foo, (void *) &foo);

This should print the same thing, since there is no variable holding the address of foo, whose address in turn can be taken.