C中的指针操作和运算符优先级 [英] Pointer operations and operator precedence in C

问题描述

今天只和一个C家伙聊天，我们不同意以下几点:

Just had a chat with a C guy today and we disagreed on the following:

int intgA[2] = { 1, 2 };
int intgB[2] = { 3, 5 };

int *intAPtr = intgA;
int *intBPtr = intgB;

所以，当我们这样做时:

So when we do:

*intAPtr++ = *intBPtr++;

我的分析

第一:

intBPtr递增1，现在指向地址5. 然后，遵从，保持值为5；

intBPtr increments by one, now pointing to the address of 5. Then, deference, holding the value 5;

intAPtr也会加一，现在指向地址2. 随后引用，值为2;

intAPtr also increments by one, now pointing to the address of 2. Subsequently referencing and the value is 2;

最后:

2替换为5.

所以它们分别是:5和5.

So respectively they are: 5 and 5.

首先将*intBPtr的值分配给*intAPtr.

因此，它们变为:3和3.

Hence, they become: 3 and 3.

然后*intAPtr和*intBPtr都加1.

因此，它们分别变为:4和4.

So, respectively they become: 4 and 4.

我认为++运算符比*和=都具有优先权，因此是我的主张.

I thought the ++ operator takes precedence both over * and =, hence my assertion.

例如，如果我们有:

*intAPtr++; 

结果应该是2，对不对?因为我们先增加指针，然后取消引用.

The result should be 2, right? Because we first increment the pointer and then dereference.

那么，在他所说的情况下，为什么在上述情况下，我们首先将intBPtr的值分配给intAPtr的值，然后最后递增这些值?

So why in the above case, as he claims, we first assign the value of intBPtr to the value of intAPtr and increment the values last?

在这里接受了所有建议之后，我在IDE中运行了代码，结果确认了@sujin的代码:

尽管它确认我至少在优先次序方面是正确的:

Although it confirms that I was right at least in terms of precedence:

那个:*intAPtr++ = *intBPtr++;

intAPtr++的优先级更高，这导致:intAPtr将其地址加1.

intAPtr++ has a higher precedence, which leads to: intAPtr increments its address by 1.

现在指向:2的地址.

同样:

intBPtr++也会增加1(地址).

intBPtr++ also increments by 1 (address).

现在指向:5的地址.

然后轮到*:

因此两者都分别被取消引用(*)到2和5.

So the both get dereferenced (*) to respectively 2 and 5.

但是问题仍然存在，因为上面的任务(=)似乎没有发生.

But the problem exists still because the assignment above (=) did not seem to take place.

如果同时执行，则两者都将变为5.

If it did both would become 5.

期待得到进一步的启发.

Looking forward to being further enlightened.

推荐答案

语句

*intAPtr++ = *intBPtr++;

解析为

*(intAPtr++) = *(intBPtr++);

并细分如下:

• intBPtr(3)当前指向的值分配给intAPtr(intgA[0])指向的位置；
• 指针 intAPtr和intBPtr递增.

• The value currently pointed to by intBPtr (3) is assigned to the location pointed to by intAPtr (intgA[0]);
• The pointers intAPtr and intBPtr are incremented.

发生这些事情的确切顺序是未指定；您不能依靠intBPtr在intAPtr之后递增，反之亦然，也不能依靠在增量之前发生的赋值，等等.

The exact order in which these things happen is unspecified; you cannot rely on intBPtr being incremented after intAPtr or vice-versa, nor can you rely on the assignment occuring before the increments, etc.

所以到intgA[0] == 3和intAPtr == &intgA[1]和intBPtr == &intgB[1]都完成了.

So by the time this is all done, intgA[0] == 3 and intAPtr == &intgA[1] and intBPtr == &intgB[1].

表达式a++计算得出的值a 之前的增量.

The expression a++ evaluates to the value of a before the increment.

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