前置与前置C#中的发布增量 [英] Pre- & Post Increment in C#

问题描述

我对C#编译器如何处理前后增量和递减感到困惑.

I am a little confused about how the C# compiler handles pre- and post increments and decrements.

当我编写以下代码时:

int x = 4;
x = x++ + ++x;

x之后将具有值10.我认为这是因为预递增将x设置为5，这使得它将5+5评估为10.然后，后增量会将x更新为6，但是不会使用此值，因为随后会将10分配给x.

x will have the value 10 afterwards. I think this is because the pre-increment sets x to 5, which makes it 5+5 which evaluates to 10. Then the post-increment will update x to 6, but this value will not be used because then 10 will be assigned to x.

但是当我编码时:

int x = 4;
x = x-- - --x;

然后x将是2.谁能解释为什么会这样?

then x will be 2 afterwards. Can anyone explain why this is the case?

推荐答案

x--将为4，但在--x时刻将为3，因此最终将为2，那么您将拥有

x-- will be 4, but will be 3 at the moment of --x, so it will end being 2, then you'll have

x = 4 - 2

顺便说一句，您的第一种情况是x = 4 + 6

btw, your first case will be x = 4 + 6

这是一个小示例，它将打印出每个零件的值，也许这样您会更好地理解它:

Here is a small example that will print out the values for each part, maybe this way you'll understand it better:

static void Main(string[] args)
{
    int x = 4;
    Console.WriteLine("x++: {0}", x++); //after this statement x = 5
    Console.WriteLine("++x: {0}", ++x); 

    int y = 4;
    Console.WriteLine("y--: {0}", y--); //after this statement y = 3
    Console.WriteLine("--y: {0}", --y);

    Console.ReadKey();
}

这将打印出来

x++: 4
++x: 6
y--: 4
--y: 2

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