c# – 当操作数为<32位时,为什么移位操作总是导致带符号int

为什么在unsigned int上的移位操作给出无符号结果,但是对较小的无符号操作数的操作会导致一个带符号的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

解决方法

shift operators仅为这些情况预定义(向左移动)：

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)

表达式uint shifUnignedShort = unsignedShort<<< 2被解释为(1)-st情况(implicit up-casting from ushort to int和(int)2),所以它对非法投射执行了一个警告(没有int结果到隐含的转换).
同样的情况我们可以看到uint shifUnsignedByte = unsignedByte<<<它还解释为(1)-st情况(从字节到int和(int)2的隐式升级,但没有将结果值隐式转换为uint). 您可以使用以下方法解决这些问题：

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