二进制转换如何设置移位操作符？

你的理解是错误的，一个转变是一个分裂或乘以2。这对于数字的二进制表示具有特殊意义。例子。10001001右移（除以2）01000 100右移。00100010（除以2）注意CPU有特殊的硬件，你的移位实际上并没有划分。现在在你的代码中，如果你向左移动，看看高位，你可以使用一个for循环，而不是8到16的比较。注意，虽然int是CPU上的16位，但并非所有CPU都是这样。INT至少为16位。



      以下为原文

    Your understanding is wrong a shift is a divide or multiply by 2. This has special meaning to a binary representation of a number.
Example.
10001001
Shift right(divide by 2)
01000100
Shift right.
00100010(divide by 2)
Note the CPU has special hardware your shift it that’s not actually divide.

Now in your code if you shift left and look at the high bit you can use a for loop rather than 8 to 16 compares.

Note while an int is 16 bits on you cpu, that is not true of all CPUs. An int is at least 16 bits.

我有一种感觉，我认为这是错误的，但是当我问我的教授她刚刚说是的时候，我知道符号char范围从-128到127，而0b100000的十进制当量是128，现在考虑int的范围从-5到255.现在对右移和除法意味着什么？通过同样的逻辑是如何应用到unsindChar，理解这些问题中的一些似乎是相当基本的，但我试图更好地掌握这个主题。



      以下为原文

   


I had a feeling I was wrong to assume that it was the case however when I asked my prof she just said yes,
 
I understand that signed char ranges from -128 to 127 and  0b10000000's decimal equivalent is 128 now to  think about int they range from -256 to 255. 
 
now what does it mean to shift right and divide by 2.
 
how is the same logic being applied to an unsigned char,understand that some of these question seem pretty basic but I am trying to grasp the subject a bit better 

我画了它，我想我有一个想法，假设这是一个0和1组成的字符列表。将其右移将意味着像从四到三的移位，这是数学除以2的等价物。



      以下为原文

    I drew it and I think I have a thought
 
_ _ _ _ _ _ _ _
 
let's assume this is a list of 0's and 1's that make up that char. shifting it right would mean something like shifting from position four to three,which is the mathematical equivalent of dividing by 2 

符号数的移位是一个更复杂的问题。C没有给出确切的定义。结果将在不同编译器之间有所不同。我说右移相当于除以2。结果对于无符号数是相同的。



      以下为原文

    Shifting of signed numbers is a more complicated subject. C does not give an exact definition of what that means. The result will vary between different compilers.

I said a shift right is the equivalent of dividing by two. The result will be the same for an unsigned number.

如果你在讨论有符号的字符，那么在XC8中，0B100000是-128，而不是+128No，一个已签名的int是16位，所以从0x7FFF＝0B01111111111111= 327 67到0x8000＝0B10000 000 000 000＝-327 68。现在，右移和除以2.0x7FFF的右移为0B011111111111111=什么意思？0x3FFF＝1638 3如果使用未签名的int，则0x8000右移变为0B0100000×0.00亿＝0x4000＝1638，但是对于签名的ItS0x8000右移变为＝0B1100000亿兆＝0xC000＝-1638，但这是特定于XC8的。C标准没有定义如果你对一个已签名的变量做了正确的转换会发生什么，而不同的编译器会做不同的事情。



      以下为原文

   
If you are talking about signed chars, then 0b10000000 is -128, not +128
 

No.
In XC8, a signed int is 16 bits, so ranges from 0x7FFF = 0b0111111111111111 = 32767
to 0x8000 = 0b1000000000000000 = -32768
 
now what does it mean to shift right and divide by 2.
0x7FFF shifted right becomes 0b0011111111111111 = 0x3FFF = 16383
 
if using unsigned ints, then
0x8000 shifted right becomes 0b0100000000000000 = 0x4000 = 16384
 
however for signed ints
0x8000 shifted right becomes= 0b1100000000000000 = 0xC000 = -16384
 
but this is specific to XC8. The C standard does NOT define what will happen if you do a right shift on a signed variable, and different compilers do it differently.
 

“考虑int的范围从256到255。”我想你会发现它稍微多一些。签署- 32768到32767或无符号0到65535。



      以下为原文

    "think about int they range from -256 to 255."
I think you'll find it slightly more than that.
 
signed -32768 to 32767 or unsigned 0 to 65535
 

亲爱的华生。



      以下为原文

    Elementary dear Watson.
 

您不需要那些& gt；& gt；and＝=，并且它将操作相同。



      以下为原文

    You don't need those >> and ==, and it will operate the same.

明确地说，Harry意味着可以简化，并且它的行为完全相同。也就是说，如果表达式的结果是非零，则将执行IF-（）语句的正文。



      以下为原文

   
Agree.
To be clear, Harry means that
if ( ((number & 0b10000000) >> 7 ) == 1)
can be simplified to
if ( number & 0b10000000 )
and it will behave exactly the same.
i.e. if the result of the expression is non-zero, the body of the if() statement will be executed.