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Round down and Round up?

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Khanh
Khanh el 30 de Dic. de 2014
Editada: DGM el 11 de Mayo de 2023
Hi everyone,
I have two numbers and want to round up and round down them.
minval = 0.4410 and maxval=0.8450
I want to round down the first number and it will be 0.4. The 2nd number after being rounded up will be 0.9.
I tried with floor(minval) it returns 0 and ceil(maxval) it returns 1. Those numbers are not the result I want.
Could someone give me some advice?
I'm using MATLAB R2014a.
Thank you so much.
  1 comentario
shaadinama
shaadinama el 8 de Mayo de 2023
Editada: DGM el 11 de Mayo de 2023
In MATLAB, the functions "floor" and "ceil" can be used to round down and round up, respectively.
The "floor" function rounds a given input value down to the nearest integer. For example, "floor(3.7)" would return 3, and "floor(-2.3)" would return -3.
The "ceil" function, on the other hand, rounds a given input value up to the nearest integer. For example, "ceil(3.2)" would return 4, and "ceil(-2.9)" would return -2.
Both functions can be useful in various mathematical and programming contexts where it is necessary to round numbers to the nearest whole integer.

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Star Strider
Star Strider el 30 de Dic. de 2014
Editada: Star Strider el 30 de Dic. de 2014
In R2014b, the round function now supports rounding to a specific number of digits:
minval = 0.4410;
maxval = 0.8450;
r1_minval = round(minval, 1);
r1_maxval = round(maxval, 1);
If you don’t have R2014b, this works:
diground = @(x,d) round(x*10^d)/10^d;
r2_minval = diground(minval,1);
r2_maxval = diground(maxval,1);
  7 comentarios
marchammer
marchammer el 25 de Feb. de 2022
Editada: marchammer el 25 de Feb. de 2022
If the requirement is based on the 2nd decimal as it could be inferred in the problem statement, given a common input and no prompt as to whether to round up or down, round(round(val,2),1) will work. Otherwise the first answer with floor() and ceil() multiplying by a power of 10 and dividing again after rounding is an easy and common way to do this.
It is interesting that the accepted answer doesn't round the second value up as requested.
Stephen23
Stephen23 el 26 de Feb. de 2022
Editada: Stephen23 el 27 de Abr. de 2023
"It is interesting that the accepted answer doesn't round the second value up as requested. "
It is simply because the actual value is less than 5 at that digit (not equal to 5 as it seems when displayed to a much lower precision):
minval = 0.4410
minval = 0.4410
maxval = 0.8450
maxval = 0.8450
fprintf('%.40f\n',minval,maxval)
0.4410000000000000031086244689504383131862 0.8449999999999999733546474089962430298328
So ROUND is working correctly, it is another reminder to always be aware of the behaviors of floating point numbers.
Given hints in the the OP's question, it seems like the correct answer was given by Rajesh here:

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Más respuestas (1)

Rajesh
Rajesh el 24 de Jul. de 2015
Movida: Stephen23 el 27 de Abr. de 2023
  1. floor(minval*10)/10
  2. ceil(maxval*10)/10
  1 comentario
Stephen23
Stephen23 el 27 de Abr. de 2023
+1 This actually returns what the OP requested:
minval = 0.4410;
maxval = 0.8450;
floor(minval*10)/10
ans = 0.4000
ceil(maxval*10)/10
ans = 0.9000

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