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

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'?

[Andrew Reibold]

Well for one, they don't equal 115 because those equal 105 and 110, haha.

I know what you mean though

[Azzi Abdelmalek]

Why 115 ?

[Jeffrey Wildman]

Oops, typo, changed 115 to 110.

[summyia qamar]

what if we want to change 10.3?what will be the format?%g is not working.

Answer 1

[per isakson]

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.

[Jeffrey Wildman]

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]

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

Answer 2

[Andrew Reibold]

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:

[Jeffrey Wildman]

Simple and effective workaround! Still kinda curious as to what is happening under the hood of sprintf.

[per isakson]

"Still kinda curious" &nbsp Don't you trust my answer?

[James Tursa]

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.

Answer 3

[Sebastian Mader]

So why did Mathworks introduce %d and %i at all? It would be safer to use %.0f in any case.

[Stephen23]

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]

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.