sprintf('%d',x) prints out exponential notation instead of decimal notation

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Jeffrey Wildman
Jeffrey Wildman 2014-8-26
评论： Sebastian Mader 2018-7-27
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'?
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per isakson
per isakson 2014-8-26
编辑：per isakson 2014-8-26
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.
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Jeffrey Wildman
Jeffrey Wildman 2014-8-26
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 2014-8-26
编辑：per isakson 2014-8-30
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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Andrew Reibold
Andrew Reibold 2014-8-26
编辑：Andrew Reibold 2014-8-26
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:
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James Tursa
James Tursa 2016-12-17
编辑：James Tursa 2016-12-17
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 2018-7-27
So why did Mathworks introduce %d and %i at all? It would be safer to use %.0f in any case.
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Stephen23
Stephen23 2018-7-27
编辑：Stephen23 2018-7-27
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 2018-7-27
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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