如何减去IEEE 754号码？ [英] How to subtract IEEE 754 numbers?

问题描述

例如：0,546875 - 32.875 ...

如何减去IEEE 754数字？在IEEE-754中$ 0111110 10001100000000000000000

- > -32.875 is 1 10000111 01000101111000000000000 in IEEE-754

那我该怎么做减法呢？我知道我必须让这两个指数平等，但是在那之后我该做什么？ 2'-32.875尾数的补充和0.546875尾数的补充？

解决方案

真的没有什么不同，你用铅笔和纸做。好的有点不同

  123400  -  5432 = 1.234 * 10 ^ 5  -  5.432 * 10 ^ 3 
  

更大的数字占主导地位，将较小数字的尾数转换为位桶，直到指数匹配为止

  1.234 * 10 ^ 5  -  0.05432 * 10 ^ 5 
  

然后执行与mantissas的减法

  1.234  -  0.05432 = 1.17968 
 1.17968 * 10 ^ 5 
  

然后标准化（在这个例子中是）

$ p
$ b

在IEEE float中，单精度

  123400 = 0x1E208 = 0b11110001000001000 
 11110001000001000.000 ... 
  

我们必须把小数点后16位移到左边，所以

  1.1110001000001000 * 2 ^ 16 
  1，并且不用单精度，我们把它除掉并保留小数部分
 
 
 
符号位，指数，尾数
 
 
  0 10001111 1110001000001000 ... 
 0100011111110001000001000 ... 
 0100 0111 1111 0001 0000 0100 0 ... 
 0x47F10400 
  
如果你编写一个程序来看看电脑的东西是什么123400是你得到同样的东西：
  0x47F10400 123400.000000 
  
所以我们知道第一个操作数的指数和尾数。 b 
 
现在第二个操作数
 
 $ p $  5432 = 0x1538 = 0b0001010100111000 
  

正常化，左移十进制12位



  1010100111000.000 
 1.010100111000000 * 2 ^ 12 
  

指数偏差加127并得到139 = 0x8B = 0b10001011

放在一起

  0 10001011 010100111000000 
 010001011010100111000000 
 0100 0101 1010 1001 1100 0000 ... 
 0x45A9C00 
  

和一个计算机程序/编译器给出相同的结果

pre code $ 0x45A9C000 5432.000000
code >

现在回答你的问题。使用浮点数的组成部分，我已经在这里恢复隐含的1，因为我们需要它

  0 10001111 111100010000010000000000  -  0 10001011 101010011100000000000000 
  

我们必须像小学一样排列小数位，然后才能减去在这种情况下，你必须把小指数移到右边，折尾数位直到指数匹配。

  0 10001111 111100010000010000000000  - 0 10001011 101010011100000000000000 
 0 10001111 111100010000010000000000  -  0 10001100 010101001110000000000000 
 0 10001111 111100010000010000000000  -  0 10001101 001010100111000000000000 
 0 10001111 111100010000010000000000  -  0 10001110 000101010011100000000000 
 0 10001111 111100010000010000000000  -  0 10001111 000010101001110000000000 
  

现在我们可以减去尾数。如果符号位匹配，那么我们将实际上相减，如果他们不匹配，然后我们添加。他们匹配这将是一个减法。

计算机通过使用加法逻辑进行减法运算，倒数第二个运算符的方式进入加法器并声明进位位这：

  1 
 111100010000010000000000 
 + 111101010110001111111111 
 ======== ================== 
  

现在只是如纸和铅笔让我们执行添加

$ pre $ $ c $ 1111000100000111111111111
111100010000010000000000
+ 111101010110001111111111
= =========================
111001100110100000000000


或者在你的计算器上用十六进制表示

$ p $ 111100010000010000000000 = 1111 0001 0000 0100 0000 0000 = 0xF10400
111101010110001111111111 = 1111 0101 0110 0011 1111 1111 = 0xF563FF
0xF10400 + 0xF563FF + 1 = 0x1E66800
1111001100110100000000000 = 1 1110 0110 0110 1000 0000 0000 = 0x1E66800

关于硬件是如何工作的，因为这实际上是一个使用加法器的减法，我们也反转了执行（或在一些电脑上，他们离开它）。所以执行1是一件好事，我们基本上放弃它。如果它是一个零，我们将需要更多的工作。我们没有执行，所以我们的答案是真正的0xE66800。

很快就可以看到另一种方式，而不是反转和添加一个让我们只用一个计算器

  111100010000010000000000  -  000010101001110000000000 = 
 0xF10400  -  0x0A9C00 = 
 0xE66800 
  

通过尝试将其可视化，我可能使情况变得更糟。尾数减法的结果是111001100110100000000000（0xE66800），在最高有效位中没有移动，在这种情况下我们以24位数字为1，没有归一化。要正常化，你需要向左或向右移动尾数，直到24位与最左边最高位1对齐，调整每个位移的指数。

现在剥离1位，我们把零件放在一起

  0 10001111 11001100110100000000000 
 01000111111001100110100000000000 
 0100 0111 1110 0110 0110 1000 0000 0000 
 0x47E66800 
  

如果您一直关注写一个程序来做到这一点，我也做到了。这个程序违反了C标准，以不正确的方式使用联合。在我的电脑上，我用编译器解决了这个问题，不要指望它一直工作。

  #include  
 union 
 {
 float f; 
 unsigned int u; 
} myun; 
 
 
 int main（void）
 {
 float a，b，c; 
 
 a = 123400; 
 b = 5432; 
 
 c = a-b; 
 
 myun.f = a; printf（0x％08X％f \ n，myun.u，myun.f）; 
 myun.f = b; printf（0x％08X％f \ n，myun.u，myun.f）; 
 myun.f = c; printf（0x％08X％f \ n，myun.u，myun.f）; 
 
 return（0）; 
} 
  

我们的结果与上面程序的输出相匹配，我们得到了一个0x47E66800
pre $ code> 0x47F10400 123400.000000
0x45A9C000 5432.000000
0x47E66800 117968.000000

所以：

<1>提取零件，符号，指数，尾数。 2）通过牺牲最小指数的数字中的尾数位来对齐小数位，将尾数移到右边，直到指数匹配为止。$ b $
$ b b
3）如果符号位相同，则执行减法运算，如果符号位不同，则执行加法运算。
$如果结果是零，那么你的答案是一个零，将IEEE值作为结果编码为零，否则：

5）将数字归一化，将答案转移到右边或左边（答案可以是24位加/减的25位，加/减可以有一个很大的正常化转换，左边或右边或许多位）直到你有一个24位的数字，其中最重要的一个左对齐。 24位用于单精度浮点数。定义规范化的更正确的方法是向左或向右移动，直到数字类似1.something。如果你有0.001，你会左移3，如果你有11.10，你会右移1，左移增加你的指数，右移减少它。 6）对于单精度去除前导1.从尾数开始，如果指数已经溢出，那么你进入建立一个信号灯。如果符号位不同，并且您执行了添加，那么您必须处理计算结果符号位。如果像以上一切都很好，你只需在结果中放置符号位，指数和尾数

乘法和除法是不同的，你问关于subract，所以这是我所覆盖的。

How do I subtract IEEE 754 numbers?

For example: 0,546875 - 32.875...

-> 0,546875 is 0 01111110 10001100000000000000000 in IEEE-754

-> -32.875 is 1 10000111 01000101111000000000000 in IEEE-754

So how do I do the subtraction? I know I have to to make both exponents equal but what do I do after that? 2'Complement of -32.875 mantissa and add with 0.546875 mantissa?

解决方案

Really not any different than you do it with pencil and paper. Okay a little different

123400 - 5432 = 1.234*10^5 - 5.432*10^3

the bigger number dominates, shift the smaller number's mantissa off into the bit bucket until the exponents match

1.234*10^5 - 0.05432*10^5

then perform the subtraction with the mantissas

1.234 - 0.05432 = 1.17968
1.17968 * 10^5

Then normalize (which in this case it is)

That was with base 10 numbers.

In IEEE float, single precision

123400 = 0x1E208 = 0b11110001000001000
11110001000001000.000...

normalize that we have to shift the decimal place 16 places to the left so

1.1110001000001000 * 2^16

The exponent is biased so we add 127 to 16 and get 143 = 0x8F. It is a positive number so the sign bit is a 0 we start to build the IEEE floating point number the leading 1 before the decimal is implied and not used in single precision, we get rid of it and keep the fraction

sign bit, exponent, mantissa

0 10001111 1110001000001000...
0100011111110001000001000...
0100 0111 1111 0001 0000 0100 0...
0x47F10400

And if you write a program to see what a computer things 123400 is you get the same thing:

0x47F10400 123400.000000

So we know the exponent and mantissa for the first operand'

Now the second operand

5432 = 0x1538 = 0b0001010100111000

Normalize, shift decimal 12 bits left

1010100111000.000
1.010100111000000 * 2^12

The exponent is biased add 127 and get 139 = 0x8B = 0b10001011

Put it all together

0 10001011 010100111000000
010001011010100111000000
0100 0101 1010 1001 1100 0000...
0x45A9C00

And a computer program/compiler gives the same

0x45A9C000 5432.000000

Now to answer your question. Using the component parts of the floating point numbers, I have restored the implied 1 here because we need it

0 10001111 111100010000010000000000 -  0 10001011 101010011100000000000000

We have to line up our decimal places just like in grade school before we can subtract so in this context you have to shift the smaller exponent number right, tossing mantissa bits off the end until the exponents match

0 10001111 111100010000010000000000 -  0 10001011 101010011100000000000000
0 10001111 111100010000010000000000 -  0 10001100 010101001110000000000000
0 10001111 111100010000010000000000 -  0 10001101 001010100111000000000000
0 10001111 111100010000010000000000 -  0 10001110 000101010011100000000000
0 10001111 111100010000010000000000 -  0 10001111 000010101001110000000000

Now we can subtract the mantissas. If the sign bits match then we are going to actually subtract if they dont match then we add. They match this will be a subtraction.

computers perform a subtraction by using addition logic, inverting the second operator on the way into the adder and asserting the carry in bit, like this:

                         1
  111100010000010000000000
+ 111101010110001111111111
==========================

And now just like with paper and pencil lets perform the add

 1111000100000111111111111
  111100010000010000000000
+ 111101010110001111111111
==========================
  111001100110100000000000 

or do it with hex on your calculator

111100010000010000000000 = 1111 0001 0000 0100 0000 0000 = 0xF10400
111101010110001111111111 = 1111 0101 0110 0011 1111 1111 = 0xF563FF
0xF10400 + 0xF563FF + 1 = 0x1E66800
1111001100110100000000000 =1 1110 0110 0110 1000 0000 0000 = 0x1E66800

A little bit about how the hardware works, since this was really a subtract using the adder we also invert the carry out bit (or on some computers they leave it as is). So that carry out of a 1 is a good thing we basically discard it. Had it been a carry out of a zero we would have needed more work. We dont have a carry out so our answer is really 0xE66800.

Very quickly lets see that another way, instead of inverting and adding one lets just use a calculator

111100010000010000000000 -  000010101001110000000000 = 
0xF10400 - 0x0A9C00 = 
0xE66800

By trying to visualize it I perhaps made it worse. The result of the mantissa subtracting is 111001100110100000000000 (0xE66800), there was no movement in the most significant bit we end up with a 24 bit number in this case with the msbit of a 1. No normalization. To normalize you need to shift the mantissa left or right until the 24 bits lines up with the most significant 1 in that left most position, adjusting the exponent for each bit shift.

Now stripping the 1. bit off the answer we put the parts together

0 10001111 11001100110100000000000
01000111111001100110100000000000
0100 0111 1110 0110 0110 1000 0000 0000
0x47E66800

If you have been following along by writing a program to do this, I did as well. This program violates the C standard by using a union in an improper way. I got away with it with my compiler on my computer, dont expect it to work all the time.

#include <stdio.h>

union
{
    float f;
    unsigned int u;
} myun;


int main ( void )
{
    float a,b,c;

    a=123400;
    b=  5432;

    c=a-b;

    myun.f=a; printf("0x%08X %f\n",myun.u,myun.f);
    myun.f=b; printf("0x%08X %f\n",myun.u,myun.f);
    myun.f=c; printf("0x%08X %f\n",myun.u,myun.f);

    return(0);
}

And our result matches the output of the above program, we got a 0x47E66800 doing it by hand

0x47F10400 123400.000000
0x45A9C000 5432.000000
0x47E66800 117968.000000

If you are writing a program to synthesize the floating point math your program can perform the subtract, you dont have to do the invert and add plus one thing, over complicates it as we saw above. If you get a negative result though you need to play with the sign bit, invert your result, then normalize.

So:

1) extract the parts, sign, exponent, mantissa.

2) Align your decimal places by sacrificing mantissa bits from the number with the smallest exponent, shift that mantissa to the right until the exponents match

3) being a subtract operation if the sign bits are the same then you perform a subtract, if the sign bits are different you perform an add of the mantissas.

4) if the result is a zero then your answer is a zero, encode the IEEE value for zero as the result, otherwise:

5) normalize the number, shift the answer to the right or left (The answer can be 25 bits from a 24 bit add/subtract, add/subtract can have a dramatic shift to normalize, either one right or many bits to the left) until you have a 24 bit number with the most significant one left justified. 24 bits is for single precision float. The more correct way to define normalizing is to shift left or right until the number resembles 1.something. if you had 0.001 you would shift left 3, if you had 11.10 you would shift right 1. a shift left increases your exponent, a shift right decreases it. No different than when we converted from integer to float above.

6) for single precision remove the leading 1. from the mantissa, if the exponent has overflowed then you get into building a signaling nan. If the sign bits were different and you performed an add, then you have to deal with figuring out the result sign bit. If as above everything fine you just place the sign bit, exponent and mantissa in the result

Multiply and divide is different, you asked about subract, so that is all I covered.

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