c - 知道初始地址的指针的地址

Question

我有以下代码:

int main(){
int tab[5]={10, 20, 30, 40, 50};
int *ptr;
ptr=tab+4;
printf("%d,%#x,%#x. \n",*ptr-1,&tab,ptr);
return 0;}

“tab”位于从地址0x28FEF8开始的内存区域。

我知道*ptr-1的值为49，&tab的值为0x28FEF8。

有人可以解释一下为什么“ptr”的值为0x28FF08吗？我猜测它会是 0x28FEFC。

提前谢谢您!

Answer 1

如果我们“绘制”您的数组，因为它在内存中布局，它将看起来像这样

+----+----+----+----+----+
| 10 | 20 | 30 | 40 | 50 |
+----+----+----+----+----+

Then you need to remember that arrays naturally decays to pointers to their first element, i.e. tab is equal to &tab[0]. And then you need to remember that for any array or pointer a and index i, the expression a[i] is exactly equal to *(a + i). From that it's easy to deduce that tab + 4 is equal to &tab[4], i.e. a pointer to the fifth element.

So if we again draw the array, but now with pointers:

+----+----+----+----+----+
| 10 | 20 | 30 | 40 | 50 |
+----+----+----+----+----+
^                   ^
|                   |
&tab[0]             &tab[4]
|
&tab

If &tab[0] is equal to 0x28fef8 then &tab[4] is equal to 0x28fef8 + sizeof(int) * 4, as it is pointing to index 4 of an int array, and it's indeed 0x28ff08.

Please note that while &tab[0] and &tab are pointing to the same location, they are semantically different. &tab[0] is a pointer to a single elements and is of type int *, while &tab is a pointer to the array and is of type int (*)[5].

---

On a related note, the correct format to print void * pointers is "%p". So your printing should really be

printf("%d,%p,%p. \n", *ptr-1, (void *) &tab, (void *) ptr);

请注意，确实需要强制转换为 void *。