为什么移位操作总是导致一个符号整数运算时为< 32位 [英] Why do shift operations always result in a signed int when operand is <32 bits

问题描述

为什么移位运算无符号整数给出一个无符号的结果，但在较小的无符号运算操作导致符号int？

  INT signedInt = 1; 
 INT shiftedSignedInt = signedInt<< 2; 
 
 UINT无符号整型= 1; 
 UINT shiftedUnsignedInt =无符号整型<< 2; //好。无符号结果
 
短signedShort = 1; 
 INT shiftedsignedShort = signedShort<< 2; 
 
 USHORT unsignedShort的= 1; 
 UINT shiftedUnsignedShort = unsignedShort的<< 2; // CS0266：不能投INT为uint 
 
为sbyte signedByte = 1; 
 INT shiftedSignedByte = signedByte<< 2; 
 
 UnsignedByte的字节= 1; 
 UINT shiftedUnsignedByte =&UnsignedByte的LT;< 2; // CS0266：不能投INT为uint 
  

解决方案

该转移只针对这些情况下预定义运算符（左移）：

  INT运营商的LT;≤（INT X，诠释计数）; （1）
 UINT运营商的LT;≤（UINT X，诠释计数）; （2）
长期运营商的LT;≤（长×，诠释计数）; （3）
 ULONG运营商的LT;≤（ULONG X，诠释计数）; （4）
  

表达式 UINT shiftedUnsignedShort = unsignedShort的<< 2 被解释为：（1）-st情况下（隐含从USHORT了铸造为int 和（INT）2 ），所以非法铸件进行警告（有从int结果没有隐式转换为USHORT）。结果
同样的情况，我们可以看到 UINT shiftedUnsignedByte =&UnsignedByte的LT;< 2 。这也解释为（1）-st情况下（隐含了铸造从byte到int和（INT）2 ，但得到的值没有隐式转换为uint）。

您可以使用以下方法解决这些问题：

  UINT shiftedUnsignedShort =（UINT）unsignedShort的<< 2 //强制使用（2）-Nd移位运算符的情况下
 UINT shiftedUnsignedByte =（UINT） -  UnsignedByte的LT;< 2; //强制使用（2）-nd移位运算符的情况下
  

Why do shift operations on unsigned ints give an unsigned result, but operations on smaller unsigned operands result in a signed int?

int signedInt = 1;
int shiftedSignedInt = signedInt << 2;

uint unsignedInt = 1;
uint shiftedUnsignedInt = unsignedInt << 2;     //OK. unsigned result

short signedShort = 1;
int shiftedsignedShort = signedShort << 2;

ushort unsignedShort = 1;
uint shiftedUnsignedShort = unsignedShort << 2; //CS0266: Can't cast int to uint

sbyte signedByte = 1;
int shiftedSignedByte = signedByte << 2;

byte unsignedByte = 1;
uint shiftedUnsignedByte = unsignedByte << 2;   //CS0266: Can't cast int to uint

解决方案

The shift operators are predefined only for these cases (shift left):

int operator <<(int x, int count);  (1)
uint operator <<(uint x, int count); (2)
long operator <<(long x, int count);  (3)
ulong operator <<(ulong x, int count); (4)

The expression uint shiftedUnsignedShort = unsignedShort << 2 is interpreted as (1)-st case (implicit up-casting from ushort to int and (int)2), so it performed a warning on illegal casting (there is no implicit cast from int result to ushort).
The same situation we can see for uint shiftedUnsignedByte = unsignedByte << 2. It also interpreted as (1)-st case (implicit up-casting from byte to int and (int)2 ,but no implicit cast of resulting value to uint).

You can resolve these issues using the following approach:

uint shiftedUnsignedShort = (uint)unsignedShort << 2  //force use the (2)-nd shift operator case  
uint shiftedUnsignedByte = (uint)unsignedByte << 2;   //force use the (2)-nd shift operator case

这篇关于为什么移位操作总是导致一个符号整数运算时为< 32位的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！