结束携带原因 [英] Reason for end around carry to do

问题描述

我知道，如果最高位带有进位，则r基数的r-1补码应该以进位结束。

I know that the r-1 complement for r-base number should do end around carry if the highest bit has carry.

但是我不知道为什么应该做到这一点。

But I cannot figure out why it should do it.

我只能想到这可能是两个表示为零的原因。

I merely can think about it is the reason may be about the two representations for zero.

ex：

 1 1 1 0    (-1)
 0 1 0 1    (+5)  
 ===============
10 0 1 1  =====>(0 1 0 0)
I just can explain it from the result that because its sum is positive, and 1's complement has two representations, so it should add one.

ex：

 1 1 1 0    (-1)
 1 0 1 0    (-5)  
 ===============
11 0 1 1  =====>(1 0 0 1)
And I cannot explain it why should add one.

真正结束结转的真正原因是什么？

What is the really reason for end around carry?

Thx，供您阅读。

推荐答案

环行进位实际上很简单：它会改变如果您认为数字是无符号的，则从r ⁿ 到r ⁿ -1的加法运算。为简化起见，让我们讨论二进制。

End-around carry is actually rather simple: it changes the modulus of the addition operation from rⁿ to rⁿ–1, if you think of the numbers as unsigned. To simplify things, let's talk about binary.

让我们使用四位二进制补码算术来计算（-2）+（-4）：

Let's compute (-2) + (-4) using four-bit two's complement arithmetic:

  1 1 1 0    (-2)
+ 1 1 0 0  + (-4)
---------  ------
1 1 0 1 0    (-6)

让进位位保持原位现在。如果将数字视为无符号整数，我们将计算出14 + 12 =26。但是，加法是对16取模，因此我们得到10，这是无符号数字，代表-6（正确的结果）。

Let's keep the carry bit where it is for now. If you look at the numbers as unsigned integers, we're computing 14 + 12 = 26. However, addition is done modulo 16, so we get 10, which is the unsigned number which represents -6 (the correct result).

在一个补码中，数字具有不同的表示形式：

In ones' complement, the numbers have different representations:

  1 1 0 1    (-2)
+ 1 0 1 1  + (-4)
---------  ------
1 1 0 0 0    (-6)

再次，让进位保持在原位。如果您将数字看成无符号整数，则我们计算的是13 + 11 =24。但是，由于回绕进位，加法运算取模15，所以我们最后得到9，它表示-6（正确

Again, let's keep the carry bit where it is. If you look at the numbers as unsigned integers, we're computing 13 + 11 = 24. However, due to the wrap-around carry, addition is done modulo 15, so we end up with 9, which represents -6 (the correct result).

所以在四位二进制补码中，-2等于14模16，-4等于12模16，而-6等于10模16。

So in four-bit two's complement, -2 is equivalent to 14 modulo 16, -4 is equivalent to 12 modulo 16, and -6 is equivalent to 10 modulo 16.

在四位补码中，-2等于13模15，-4等于11模15，而-6是等于9模15。

And in four-bit ones' complement, -2 is equivalent to 13 modulo 15, -4 is equivalent to 11 modulo 15, and -6 is equivalent to 9 modulo 15.

有符号零：之所以得到有符号零是因为四位中有16个可能的数字，但是如果您要进行15模运算，则0和15是等效的。就是这么多。

Signed zero: The reason you get "signed zero" is because there are 16 possible numbers in four bit, but if you're doing modulo-15 arithmetic, then 0 and 15 are equivalent. That's all there is to it.

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