How to improve double precision?

7 次查看（过去 30 天）
NoYeah
NoYeah 2020-7-23
编辑： Stephen23 2020-7-25
I have made below code
a=11;
b=256;
c=a/b
and the output said c=0, which was not I had expected
So I have used cast operator
a=11;
b=256;
c=double(a/b)
and the result returns the same value; 0
I have searched Matlab document and found that the default precision of double data type is 32
so It should return 0.04296875 instead of 0. But still get 0 for this kind of codes
How to fix it properly?
  4 个评论
显示 2更早的评论
dpb
dpb 2020-7-23
I asked you to type in teh above commands on your system and tell us what you get?
Stephen23
Stephen23 2020-7-23
编辑：Stephen23 2020-7-25
"I have searched Matlab document and found that the default precision of double data type is 32"
Highly unlikely. Can you please provide a link to the MATLAB documentation where this value comes from?

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Walter Roberson
Walter Roberson 2020-7-23
Either your a or your b is type uint8. When you work with int8, int16, int32, uint8, uint16, or uint32, then the operands are converted to double(), and the operation is done, and then the result is converted back to the most restricted type used.
Operations for int64 and uint64 are handled differently in some, because 64 bit integers cannot be converted to double without losing precision.
  1 个评论
dpb
dpb 2020-7-23
编辑：dpb 2020-7-23
Could be any of the int classes, of course...what the
whos a b c
would have told us is which one(s) OP had...what OP didn't show was how a, b were created to begin with to become int class

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Logical 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by