d****n
发帖数: 130 | 1 int a[] = {3, 2};
cout << *(a+1) << endl;
cout << *(&a+1);
&a应该是变量a的地址吧？a和&a应该不是一回事吧？ |
n******t
发帖数: 4406 | 2 In C/C++, the name of a array in an expression decays to a pointer
to its first element EXCEPT it is the operand of & or sizeof.
【在 d****n 的大作中提到】
: int a[] = {3, 2};
: cout << *(a+1) << endl;
: cout << *(&a+1);
: &a应该是变量a的地址吧？a和&a应该不是一回事吧？
|
l*****d
发帖数: 359 | 3 如果a是数组， a就是&a吧，一回事儿。
【在 d****n 的大作中提到】
: int a[] = {3, 2};
: cout << *(a+1) << endl;
: cout << *(&a+1);
: &a应该是变量a的地址吧？a和&a应该不是一回事吧？
|
f**y
发帖数: 138 | 4 Here a and &a are both pointers and have the same value.
But a + 1 and &a + 1 are different. a is a simple pointer, &a is pointer of
pointer.
a + 1 points to the next element in the array.
&a + 1 points to the address after a, so &a + 1 == (void *)a + sizeof(a) |
l*****d
发帖数: 359 | 5 it seems that a is a pointer to an element, while &a is a pointer to an
array of elements
of
【在 f**y 的大作中提到】
: Here a and &a are both pointers and have the same value.
: But a + 1 and &a + 1 are different. a is a simple pointer, &a is pointer of
: pointer.
: a + 1 points to the next element in the array.
: &a + 1 points to the address after a, so &a + 1 == (void *)a + sizeof(a)
|
