整数提升 - 什么是步骤 [英] Integer promotion - what are the steps

问题描述

此代码打印B2

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

我读到整数提升，但它仍然不清楚，它在这里的例子是如何工作的？有人可以彻底地发布编译器在加宽/截断值时所遵循的步骤吗？

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

>

Let's walk through your code:

short a = -5;

a = -5，到目前为止很容易。

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

unsigned short b = -5u;

-5u表示应用一元 - 到常数5u。 5u是（unsigned int）5，并且一元 - 没有促销，所以你最终得到4294967291这是2 ^ 32-5。 （更新：我的原始答案有点错误;看到一个测试脚本显示此版本是正确的在这里 http://codepad.org/hjooaQFW ）

-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)

现在当把它放在b中时，它被截断为一个无符号short（通常为2个字节），所以b = 65531，即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 )

在这一行中，a和b都被提升为int，以便比较可以正确进行。如果他们被晋升为短裤，b可能缠绕。如果他们被提升为未签名的短裤，一个可能的环绕。

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.

这就像说 if（（int）a == b）。和a = -5，所以（int）a = -5，b = 65531，所以（int）b = 65531，因为int比shorts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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