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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?
When
tis 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
tis used as the argument of the&operator, no such conversion takes place. The&then explicitly takes the address oft(the array).&tis 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.