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 Comments

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Accepted Answer

per isakson
per isakson on 26 Aug 2014
Edited: per isakson on 26 Aug 2014
What you see is a consequence of how floating point arithmetic works.
See:
1.05*100 evaluates to a whole number (flint). The other two don't.
  2 Comments
Jeffrey Wildman
Jeffrey Wildman on 26 Aug 2014
Thanks for the links. I was aware of floating point representation/arithmetic, but had assumed MATLAB would perform some implicit conversion of types from float to int based on the conversion type specified in sprintf. This assumption must be incorrect.
per isakson
per isakson on 26 Aug 2014
Edited: per isakson on 30 Aug 2014
Somewhere down the page fprintf, Write data to text file it says:
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

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More Answers (2)

Andrew Reibold
Andrew Reibold on 26 Aug 2014
Edited: Andrew Reibold on 26 Aug 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 Comments
James Tursa
James Tursa on 17 Dec 2016
Edited: James Tursa on 17 Dec 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.

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Sebastian Mader
Sebastian Mader on 27 Jul 2018
So why did Mathworks introduce %d and %i at all? It would be safer to use %.0f in any case.
  2 Comments
Stephen23
Stephen23 on 27 Jul 2018
Edited: Stephen23 on 27 Jul 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
Sebastian Mader on 27 Jul 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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