整型提升 - 哪些步骤

问题描述

这code打印B2

short a=-5;
unsigned short b=-5u;
if(a==b)
    printf("A1");
else
    printf("B2");

我读到关于整型提升，但目前还不清楚对我来说，它是如何在这里的例子工作？有人能彻底发布的编译器遵循加宽的步骤/截断值？

I read about integer promotion but it's still unclear to me, how does it work in the example here? Can someone thoroughly post the steps the compiler follows in widening/truncating the values?

推荐答案

让我们通过您的code：

Let's walk through your code:

short a = -5;

A = -5，它适合短。到目前为止那么容易了。

a = -5, which fits into a short. So far so easy.

unsigned short b = -5u;

-5u指应用一元 - 运营商不断5U。 5U是（无符号整数）5，和一元 - 确实没有推广，让你最终4294967291是2 ^ 32-5。 （更新：我得到这个位错在我原来的答复;看到一个测试脚本，显示该版本是在这里正确）

-5u means apply the unary - operator to the constant 5u. 5u is (unsigned int) 5, and the unary - does no promotion, so you end up with 4294967291 which is 2^32-5. (Update: I got this bit wrong in my original answer; see a test script which shows this version is correct here http://codepad.org/hjooaQFW)

现在将仅在B时，被截断的无符号短（2字节，通常情况下），所以B = 65531，这是2 ^ 16-5。

Now when putting that in b, it is truncated to an unsigned short (2 bytes, usually), so b = 65531, which is 2^16-5.

if( a == b )

在这条线，a和b都被提升到整数，使得比较可以正确地发生。如果他们被提拔到短路，B将可能环绕。如果他们被提拔到无符号短裤，将有可能缠绕。

In this line, a and b are both promoted to ints so that the comparison can happen correctly. If they were promoted to shorts, b would potentially wrap around. If they were promoted to unsigned shorts, a would potentially wrap around.

因此​​，它好像是说 IF（（INT）一==（INT）B）。且a = -5，所以（int）的一个= -5，以及b = 65531，所以（int）的B = 65531，因为整数比短裤更大

So it's like saying if( (int) a == (int) b ). And a = -5, so (int) a = -5, and b = 65531, so (int) b = 65531, because ints are bigger than shorts.

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