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:
The actual type of t is int[10], so &t is the address of the array.
Also, int[] implicitly converts to int*, so t converts to the address of the first element of the 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.
来源：https://stackoverflow.com/questions/8412694/address-of-an-array