我只是好奇,为了在二进制中表示-1,是否有一个原因,使用了2的补码:翻转位并加1?

-1由11111111(2的补码)表示,而不是(对我来说更直观)10000001,10000001是二进制1,第一位为负标志。

免责声明:我的工作不依赖二进制算术!


当前回答

2的补数允许以正常的方式进行加减法(就像对无符号数字进行绕线运算一样)。它还防止了-0(一种单独的表示0的方法,如果使用常规的逐位比较数字的方法,它将不等于0)。

其他回答

2的补数允许以正常的方式进行加减法(就像对无符号数字进行绕线运算一样)。它还防止了-0(一种单独的表示0的方法,如果使用常规的逐位比较数字的方法,它将不等于0)。

A major advantage of two's-complement representation which hasn't yet been mentioned here is that the lower bits of a two's-complement sum, difference, or product are dependent only upon the corresponding bits of the operands. The reason that the 8 bit signed value for -1 is 11111111 is that subtracting any integer whose lowest 8 bits are 00000001 from any other integer whose lowest 8 bits are 0000000 will yield an integer whose lowest 8 bits are 11111111. Mathematically, the value -1 would be an infinite string of 1's, but all values within the range of a particular integer type will either be all 1's or all 0's past a certain point, so it's convenient for computers to "sign-extend" the most significant bit of a number as though it represented an infinite number of 1's or 0's.

Two's-complement is just about the only signed-number representation that works well when dealing with types larger than a binary machine's natural word size, since when performing addition or subtraction, code can fetch the lowest chunk of each operand, compute the lowest chunk of the result, and store that, then load the next chunk of each operand, compute the next chunk of the result, and store that, etc. Thus, even a processor which requires all additions and subtractions to go through a single 8-bit register can handle 32-bit signed numbers reasonably efficiently (slower than with a 32-bit register, of course, but still workable).

当使用C标准所允许的任何其他有符号表示时,结果的每一位都可能受到操作数的任何位的影响,这就需要将整个值一次保存在寄存器中,或者在计算之后进行额外的步骤,至少在某些情况下,需要读取、修改和重写结果的每个块。

这样加法就不需要任何特殊的逻辑来处理负数了。在维基百科上查看这篇文章。

假设有两个数,2和-1。在表示数字的“直观”方式中,它们将分别为0010和1001(我坚持使用4位的大小)。两者互为补足,分别是0010和1111。现在,假设我想把它们相加。

2的补语加法非常简单。你通常加数字,任何进位在最后被丢弃。所以它们相加如下:

  0010
+ 1111
=10001
= 0001 (discard the carry)

0001是1,这是“2+(-1)”的预期结果。

但在你的“直观”方法中,添加更复杂:

  0010
+ 1001
= 1011

等于-3,对吧?简单的加法在这种情况下行不通。你需要注意,其中一个数字是负的,如果是这种情况,就使用不同的算法。

对于这种“直观的”存储方法,减法是一种不同于加法的操作,在加法之前需要对数字进行额外的检查。由于您希望最基本的操作(加法、减法等)尽可能快,因此需要以允许您使用尽可能简单的算法的方式存储数字。

此外,在“直观”存储方法中,有两个0:

0000  "zero"
1000  "negative zero"

它们直观上是相同的数字,但存储时有两个不同的值。每个应用程序都需要采取额外的步骤来确保非零值也不是负零。

以这种方式存储int型还有另一个好处,那就是当你需要扩展存储值的寄存器的宽度时。对于2的补数,在8位寄存器中存储一个4位数就是重复它的最高位:

    0001 (one, in four bits)
00000001 (one, in eight bits)
    1110 (negative two, in four bits)
11111110 (negative two, in eight bits)

这只是观察小单词的符号位,然后重复它,直到它赶上大单词的宽度。

使用你的方法,你需要清除现有的位,这是一个额外的操作,除了填充:

    0001 (one, in four bits)
00000001 (one, in eight bits)
    1010 (negative two, in four bits)
10000010 (negative two, in eight bits)

在这两种情况下,您仍然需要设置额外的4位,但在“直观”情况下,您还需要清除第5位。这是每个应用程序中最基本和最常见操作之一中的一个小小的额外步骤。

It's worthwhile to note that on some early adding machines, before the days of digital computers, subtraction would be performed by having the operator enter values using a different colored set of legends on each key (so each key would enter nine minus the number to be subtracted), and press a special button would would assume a carry into a calculation. Thus, on a six-digit machine, to subtract 1234 from a value, the operator would hit keys that would normally indicate "998,765" and hit a button to add that value plus one to the calculation in progress. Two's complement arithmetic is simply the binary equivalent of that earlier "ten's-complement" arithmetic.

使用2的补码是因为它更容易在电路中实现,也不允许负零。

如果有x位,2的补码范围从+(2^x/2+1)到-(2^x/2)。补码将从+(2^x/2)到-(2^x/2),但允许负数为零(0000在4位1的补码系统中等于1000)。