sprintf('%d',x) prints out exponential notation instead of decimal notation
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I am using version '8.3.0.532 (R2014a)'. The sprintf command seems to print out exponential notation when decimal notation is requested (second and third example):
sprintf('%d',1.05*100)
sprintf('%d',1.10*100)
sprintf('%.0d',1.10*100)
ans = 105
ans = 1.100000e+02
ans = 1e+02
Is there any reason why the last two calls are not printing '110'?
4 件のコメント
summyia qamar
2016 年 12 月 16 日
what if we want to change 10.3?what will be the format?%g is not working.
採用された回答
per isakson
2014 年 8 月 26 日
編集済み: per isakson
2014 年 8 月 26 日
What you see is a consequence of how floating point arithmetic works.
See:
- http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F and read
- Cleve's piece
1.05*100 evaluates to a whole number (flint). The other two don't.
2 件のコメント
per isakson
2014 年 8 月 26 日
編集済み: per isakson
2014 年 8 月 30 日
If you specify a conversion that does not fit the data, such as
a string conversion for a numeric value, MATLAB overrides the
specified conversion, and uses %e.
To me this was "expected behavior", but I had to look it up now. One cannot read and remember everything. Thus, when in doubt make a test
>> sprintf( '%d', 1/3 )
ans =
3.333333e-01
その他の回答 (2 件)
Andrew Reibold
2014 年 8 月 26 日
編集済み: Andrew Reibold
2014 年 8 月 26 日
Use f instead of d for floating point notation will stop the scientific I believe.
sprintf('%f',1.05*100)
sprintf('%f',1.10*100)
sprintf('%.0f',1.10*100)
ans = 105.000000
ans = 110.000000
ans = 110
Notice I can stop the decimals by using .0f like I did in the last example.
For additional reference:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/175465/image.png)
3 件のコメント
James Tursa
2016 年 12 月 17 日
編集済み: James Tursa
2016 年 12 月 17 日
This is what is happening "under the hood" with the floating point numbers (neither 1.05 nor 1.10 can be represented exactly in IEEE double):
>> num2strexact(1.05)
ans =
1.0500000000000000444089209850062616169452667236328125
>> num2strexact(1.05*100)
ans =
1.05e2
>> num2strexact(1.10)
ans =
1.100000000000000088817841970012523233890533447265625
>> num2strexact(1.10*100)
ans =
1.100000000000000142108547152020037174224853515625e2
You got lucky on the 1.05*100 that it resulted in 105 exactly, but you didn't get lucky in the 1.10*100 case.
Sebastian Mader
2018 年 7 月 27 日
So why did Mathworks introduce %d and %i at all? It would be safer to use %.0f in any case.
2 件のコメント
Stephen23
2018 年 7 月 27 日
編集済み: Stephen23
2018 年 7 月 27 日
They are not the same thing at all! For integer types, %u, %d and %i formats give the full precision, whereas what you propose does not:
>> sprintf('%.0f',intmax('uint64')) % rounded
ans =
18446744073709552000
>> sprintf('%u',intmax('uint64')) % full precision
ans =
18446744073709551615
>> sprintf('%.0f',intmax('int64')) % rounded
ans =
9223372036854775800
>> sprintf('%i',intmax('int64')) % full precision
ans =
9223372036854775807
It is obvious from the number of output digits that the '%f' format performs rounding operations using double class.
Sebastian Mader
2018 年 7 月 27 日
I see your Point, thanks for being very clary on this, much appreciated. I am far from the Limits, where rounding becomes an issue with '%.0f', so I can savely use this approach.
Nonetheless, I believe that the comments on "Notable Behavior of Conversions with Formatting Operators" should be moved up in the documentation and the special case of using %d with double precison numbers mentioned. It is at least to me not obvious at all, that an implicit type conversion is not performed by fprintf despite my desire to print an integer.
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