VBA 舍入问题 [英] VBA rounding problem

问题描述

我在 VBA 中有这个晦涩的舍入问题.

I have this obscure rounding problem in VBA.

a = 61048.4599674847
b = 154553063.208822
c = a + b   
debug.print c
Result: 
154614111.66879 

问题来了，为什么 VBA 将变量 c 舍入?我没有发出任何舍入功能.我期望的值是 154614111.6687894847.即使我四舍五入或将变量 c 格式化为小数点后 15 位，我仍然没有得到预期的结果.

Here is the question, why did VBA rounded off variable c? I didn't issued any rounding off function. The value I was expecting was 154614111.6687894847. Even if I round off or format variable c to 15 decimal places I still don't get my expected result.

任何解释将不胜感激.

使用 cDec 获得了预期的结果.我在 Jonathan Allen 在 为什么 CLng 产生不同的结果?

Got the expected results using cDec. I have read this in Jonathan Allen's reply in Why does CLng produce different results?

测试结果如下:

a = cDec(61048.4599674847)
b = cDec(154553063.208822)
c = a + b
?c
154614111.6687894847 

推荐答案

原因是可以存储在浮点变量中的精度有限.
要获得完整的解释，您应该阅读 David Goldberg 于 1991 年 3 月出版的 Computing Surveys 论文What Every Computer Scientist should Know About Floating-Point Arithmetic.

The reason is the limited precission that can be stored in a floating point variable.
For a complete explanation you shoud read the paper What Every Computer Scientist Should Know About Floating-Point Arithmetic, by David Goldberg, published in the March, 1991 issue of Computing Surveys.

论文链接

在 VBA 中，默认浮点类型是 Double，这是一个 IEEE 64 位(8 字节)浮点数.

In VBA the default floating point type is Double which is a IEEE 64-bit (8-byte) floating-point number.

还有另一种类型:Decimal，它是一个 96 位(12 字节)有符号整数，按 10 的可变幂进行缩放
简而言之，这提供了 28 位精度的浮点数.

There is another type available: Decimal which is a 96-bit (12-byte) signed integers scaled by a variable power of 10
Put simply, this provides floating point numbers to 28 digit precission.

在您的示例中使用:

a = CDec(61048.4599674847)
b = CDec(154553063.208822)
c = a + b   
debug.print c
Result: 
154614111.6687894847 

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