sprintf('%d',x) prints out exponential notation instead of decimal notation
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Jeffrey Wildman
el 26 de Ag. de 2014
Comentada: Sebastian Mader
el 27 de Jul. de 2018
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 comentarios
summyia qamar
el 16 de Dic. de 2016
what if we want to change 10.3?what will be the format?%g is not working.
Respuesta aceptada
per isakson
el 26 de Ag. de 2014
Editada: per isakson
el 26 de Ag. de 2014
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 comentarios
per isakson
el 26 de Ag. de 2014
Editada: per isakson
el 30 de Ag. de 2014
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
Más respuestas (2)
Andrew Reibold
el 26 de Ag. de 2014
Editada: Andrew Reibold
el 26 de Ag. de 2014
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:
3 comentarios
James Tursa
el 17 de Dic. de 2016
Editada: James Tursa
el 17 de Dic. de 2016
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
el 27 de Jul. de 2018
So why did Mathworks introduce %d and %i at all? It would be safer to use %.0f in any case.
2 comentarios
Stephen23
el 27 de Jul. de 2018
Editada: Stephen23
el 27 de Jul. de 2018
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
el 27 de Jul. de 2018
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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