是什么(int_1 + = * pointer ++ = int_2 ++)< int_3是什么意思? [英] What does (int_1 += *pointer++ = int_2++) < int_3 mean?

问题描述

我正在阅读这个较早的答案，其中包含一段我无法理解的C代码.基本上看起来像这样:

I was reading this earlier answer which had a piece of C code I couldn't understand. It essentially looks like this:

if((int_1 += *pointer++ = int_2++) < int_3) continue;

我可以将其分解为这样的内容-

I can break it down to something like this -

这是什么意思?我可以起床了:

What does this mean? I can get up to about this point:

if((int_1 = int_1+ *pointer++ (unsure about this part))<int_3) continue;

推荐答案

因此对于初学者来说，这确实是非常糟糕的C代码.就像可怕的C代码一样.就像，我已经用C编码很长时间了，不得不使用 operator优先级图表，因为我从未遇到过如此糟糕的情况.因此，没有理由编写这样的东西-当然不在生产代码中，并且希望不作为类的一部分，因为您永远都不需要知道操作员优先级的这个特殊问题(来源:我以谋生为生).我要说的是，您所引用的源代码是不应该以这种方式编写的错误的C代码.

So for starters, this is really, really bad C code. Like, horrible C code. Like, I've been coding in C for a long time and had to pull up an operator precedence chart because I've never encountered something like this terrible. So there's no reason to write something like this - certainly not in production code, and hopefully not as part of a class because you shouldn't never need to know this particular quirk of operator precedence (source: I teach CS for a living). I would go so far as to say that the source that you're referencing is Bad C Code That Should Never Be Written That Way.

但是话虽这么说，让我们分解一下！这里的核心表达是

But with that said, let's break it down! The core expression here is

(int_1 += *pointer++ = int_2++) < int_3

在该内部表达式中有两个赋值运算符，它们具有相同的优先级，并且从右到左具有相同的组.这意味着等同于

In that inner expression are two assignment operators, which have equal precedence and group from right-to-left. That means this is equivalent to

(int_1 += (*pointer++ = int_2++)) < int_3

这是

1. 增加int_2并存储其旧值.
2. 将该值存储在pointer所指向的位置，然后将指针前进到下一个位置.
3. 将该值添加到int_1.
4. 然后查看该值是否小于int_3.

1. Increment int_2 and store its old value.
2. Store that value at the place pointed at by pointer, then advance pointer to the next location.
3. Add that value to int_1.
4. Then see whether that value is less than int_3.

没有理由做这样的事情.只需写

There's no reason to do something like this. Just write

int val = int_2;
*pointer = val;
int_1 += val;

int_2++;
pointer++;

if (int_1 < int_3) {
    ...
}

是的！不要写这样的代码.曾经. :-)

So yeah! Don't write code like this. Ever. :-)

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