Question

*（P1 ++）

int array[10] = {1,2};
int *p1 = array;
*p1=24;
*p1= *(p1++);
for (int i : array)
    cout << i << " ";

输出为24 24

*（++ P1）

int array[10] = {1,2};
int *p1 = array;
*p1=24;
*p1= *(++p1);
for (int i : array)
    cout << i << " ";

输出为24 2

看起来这与使用值增量完全相反。有人能解释一下这里发生了什么吗？谢谢！

Answer 1

There is an undefined behavior in

*p1 = *(p1++);

because, quoting §1.9/15:

If a side effect on a scalar object is unsequenced relative to either another side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined.

Side effect here is an increment of p1 and value computation is a computation of address using p1.

So you shouldn't rely on the exact outcome of your examples.

Answer 2

*p1= *(p1++);

This just doesn't make sense. The semantic meaning of this operation is different depending on which side of the = is evaluated first. So there's no way you can make any sense out of it.

Answer 3

对于*（p1 ++）：

*p1 = *(p1++)

p1++会将p1增加到指向数组中的索引1，并返回先前的p1值（索引0）。因此*(p1++)将返回24，*p1现在将等于2. *p1然后被指定返回值（24），因此数组将为{24,24}。

for *（++ p1）：

*p1 = *(++p1)

++p1会将p1增加为指向数组中的索引2，并返回当前值p1（索引1）。因此*(++p1)将返回2，*p1现在将等于2. *p1然后被赋予返回值（2），这是索引为p1（1）的原始值，所以数组将保持{24,2}

C ++中的前/后增量指针

3 个答案: