フィルターのクリア

When with rounding and precision

1 回表示 (過去 30 日間)
Daulton_Benesuave
Daulton_Benesuave 2016 年 2 月 18 日
コメント済み: MHN 2016 年 2 月 19 日
I am calculating the following: p15 = 0.8683*1.15
When I round to 5 decimal places round(p15,5) the answer is 0.99854 since Matlab sees full precision as 0.9985449999999999.
You can confirm this by typing sprintf('%1.30f',.8683*1.15)
ans = 0.998544999999999900000000000000
Unfortunately, I need to match the output of a standard calculator (don't ask why) and need to have the output as 0.99855. How is this possible via Matlab?

採用された回答

Walter Roberson
Walter Roberson 2016 年 2 月 19 日
round(p15 * 1e6)/1e6
By the way, you cannot see the full precision on MS Windows by using sprintf(), but you can on OS-X
0.99854499999999990489385481851059012115001678466796875

その他の回答 (1 件)

MHN
MHN 2016 年 2 月 18 日
What you said is not true! for multiplication the exact number is : 0.998545000000000
p15 = 0.8683*1.15;
r = round(p15,5);
then p15 = 0.998545000000000 and r = 0.998550000000000.
  5 件のコメント
Walter Roberson
Walter Roberson 2016 年 2 月 19 日
The MS Windows version of the C library is rubbish at printing out complete numbers. The Linux version is better now but there was historically a period during which the library used for it had an error in extended printing. I do not know of the OS-X library has always been correct, but I do not recall hearing of any problems for it in this regard.
MHN
MHN 2016 年 2 月 19 日
I did not mean that there is a "problem" in sprintf. I meant even when we just assign a value like a=0.1; and then we use sprintf('%1.50f',a) the exact value is not 0.1. So, as long as one use sprintf and the floating point system, it will not get the result 0.100000000000000000000000000000000000000000.

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeAxis Labels についてさらに検索

タグ

Community Treasure Hunt

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

Start Hunting!

Translated by