为什么这两个指针减法给不同的结果？ [英] Why do these two pointer subtractions give different results?

问题描述

请考虑以下代码：

char* p = new char[2];
long* pi = (long*) p;
assert(p == pi);         // OK

char* p1 = &p[1];
long* pi1 = (long*) p1;
assert(p1 == pi1);       // OK

int d = p1 - p;
int d1 = pi1 - pi;
assert(d == d1);         // No :(

运行后，我得到 d == 1 和 d1 == 0 ，但 p1 == pi1 > p == pi （我在调试器中检查过）。这是未定义的行为吗？

After this runs, I get d == 1 and d1 == 0, although p1 == pi1 and p == pi (I checked this in the debugger). Is this undefined behavior?

推荐答案

指针之间的区别是元素的数量，不是它们之间的字节数。

The difference between pointers is the number of elements, not the number of bytes between them.

pi和pi1都指向long，但是pi1指向的地址比pi只是一个字节，假设longs是4字节长，地址的差异，1除以元素的大小，4，是0.

pi and pi1 both point to longs, but the address pointed to by pi1 is only one byte further than pi. Presuming longs are 4 bytes long, the difference in the addresses, 1, divided by the size of the element, 4, is 0.

另一种思考方式是，你可以想象编译器将生成与此相等的代码以计算d1：

Another way of thinking of this is you could imagine the compiler would generate code equivalent to this for calculating d1:

int d1 = ((BYTE*)pi1 - (BYTE*)pi)/sizeof(long).

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