When with rounding and precision

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Daulton_Benesuave
Daulton_Benesuave el 18 de Feb. de 2016
Comentada: MHN el 19 de Feb. de 2016
I am calculating the following: p15 = 0.8683*1.15
When I round to 5 decimal places round(p15,5) the answer is 0.99854 since Matlab sees full precision as 0.9985449999999999.
You can confirm this by typing sprintf('%1.30f',.8683*1.15)
ans = 0.998544999999999900000000000000
Unfortunately, I need to match the output of a standard calculator (don't ask why) and need to have the output as 0.99855. How is this possible via Matlab?

Respuesta aceptada

Walter Roberson
Walter Roberson el 19 de Feb. de 2016
round(p15 * 1e6)/1e6
By the way, you cannot see the full precision on MS Windows by using sprintf(), but you can on OS-X
0.99854499999999990489385481851059012115001678466796875

Más respuestas (1)

MHN
MHN el 18 de Feb. de 2016
What you said is not true! for multiplication the exact number is : 0.998545000000000
p15 = 0.8683*1.15;
r = round(p15,5);
then p15 = 0.998545000000000 and r = 0.998550000000000.
  5 comentarios
Walter Roberson
Walter Roberson el 19 de Feb. de 2016
The MS Windows version of the C library is rubbish at printing out complete numbers. The Linux version is better now but there was historically a period during which the library used for it had an error in extended printing. I do not know of the OS-X library has always been correct, but I do not recall hearing of any problems for it in this regard.
MHN
MHN el 19 de Feb. de 2016
I did not mean that there is a "problem" in sprintf. I meant even when we just assign a value like a=0.1; and then we use sprintf('%1.50f',a) the exact value is not 0.1. So, as long as one use sprintf and the floating point system, it will not get the result 0.100000000000000000000000000000000000000000.

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