C-具有前后增量的指针行为 [英] C - pointer behavior with pre and post increment
问题描述
我正在用C指针做一些实验，试图理解它的行为.以下是我对以下代码的假设.如果我错了，请纠正我. 我有以下代码:
In am doing some experiment in C pointers and trying to understand its behaviour. The following are my assumptions for the below codes. Correct me if I am wrong. I have the following codes:
int n[5] = {3,6,9,12,15};
int *ptr = n;
ptr++;
printf("%d", *ptr); //Needless to say, the output will be 6
我的假设:上面的输出为6，因为ptr++表示ptr = ptr + 1我正在更改ptr的值，即地址在n[0].
My assumption: The output above is 6 because ptr++ means ptr = ptr + 1 I am changing the value of ptr which is the address of n[0].
现在，我们来看看以下情形:
Now we take a look at the following scenario:
int n[5] = {3,6,9,12,15};
int *ptr = n;
*ptr++;
printf("%d", *ptr); //Why is the output still 6?
我的问题是:我们如何解释*ptr++?这是否意味着:
My question is: How do we interpret *ptr++? Does it means:
- * ptr = * ptr +1或
- * ptr = ptr + 1或
- ptr = ptr + 1或什么?
顺便说一句，当我打印出n的值时，它仍然是3,6,9,12,15.
By the way, when I print out values of n, it is still 3,6,9,12,15.
对不起，第二个问题:
那我们该如何解释*++ptr和++*ptr?
推荐答案
表达式
*ptr++;
具有价值
*ptr
在递增ptr之前，然后ptr本身就是递增.没有写意
before incrementing ptr and then ptr is increment itself. There is no sense to write
*ptr++;
因为未使用在ptr递增之前的*ptr表达式值.所以实际上这些表达式(表达式语句)的结果
because the value of expression that is *ptr before incrementing of ptr is not used. So in fact the result of these expressions (expression-statements)
ptr++;
和
*ptr++;
是相同的.
对于表达式*++ptr和++*ptr，则在此表达式中，第一个ptr的* ++ ptr递增(即它将指向数组的第二个元素)，然后被取消引用，其值是第二个要素.
在此表达式++*ptr中，首先返回数组的第一个元素的值(即3)，然后将该值递增，您将得到4.
As for expressions *++ptr and ++*ptr then in this expression *++ptr at first ptr is incremented ( that is it will point to the second element of the array) and then dereferenced and its value is the value of the second element.
In this expression ++*ptr at first the value of the first element of the array is returned (that is 3) and then this value is incremented and you will get 4.
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