为什么CLng产生不同的结果？ [英] Why does CLng produce different results?

问题描述

这是我的VBE（MS Excel 2007 VBA）一个小宝石：

 ？clng（150 * 0.85）
 127 
x = 150 * 0.85 
？clng（x）
 128 
  

任何人都可以解释这个行为吗？ IMHO第一个表达式应该产生128（.5舍入到最接近的偶数），或者至少应该两个结果相等。

解决方案

在语句 clng（150 * 0.85）， 150 * 0.85 以扩展精度计算：

  150 = 1.001011 x 2 ^ 7 
  

0.85 double precision =

  1.1011001100110011001100110011001100110011001100110011 x 2 ^ -1 
  

手动将这些数字相乘，即可得到

  1.1111110111111111111111111111111111111111111111111111111001 x 2 ^ 6 = 
 127.4999999999999966693309261245303787291041245303787291049957275390625 
  

这是59位，适合扩展精度。 $ x = 150 * 0.85 127.5 ，该59位值舍入为53位，给出

  1.1111111 x 2 ^ 6 = 1111111.1 = 127.5 
  

因此它根据round-half-to-even舍入。

（请参阅我的文章 http： //www.exploringbinary.com/when-doubles-dont-behave-like-doubles/ 了解详情。）

Here's a little gem directly from my VBE (MS Excel 2007 VBA):

?clng(150*0.85)
 127 
x = 150*0.85
?clng(x)
 128 

Can anybody explain this behaviour? IMHO the first expression should yield 128 (.5 rounded to nearest even), or at least should both results be equal.

解决方案

I think wqw is right, but I'll give the details.

In the statement clng(150 * 0.85), 150 * 0.85 is calculated in extended-precision:

150 = 1.001011 x 2^7

0.85 in double precision =

1.1011001100110011001100110011001100110011001100110011 x 2^-1

Multiply these by hand and you get

1.1111110111111111111111111111111111111111111111111111110001 x 2^6 =
127.4999999999999966693309261245303787291049957275390625

That's 59 bits, which fits comfortably in extended-precision. It's less than 127.5 so rounds down.

In the statement x = 150 * 0.85, that 59 bit value is rounded to 53 bits, giving

1.1111111 x 2^6 = 1111111.1 = 127.5

So it rounds up according to round-half-to-even.

(See my article http://www.exploringbinary.com/when-doubles-dont-behave-like-doubles/ for more information.)

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