Slider changes Pi to a MachinePrecision number

Question

Problem

Consider the following code:

DynamicModule[{a = {0, Pi}, b = 4},
  {Slider[Dynamic@b, {a}], Dynamic@b}
]

I want a Slider which sets b to the exact values in a (0 and Pi). However b is set to 3.14159 instead of Pi.

The same problem exists also for Manipulator and Manipulate:

Manipulate[b,
  {b, {0, Pi}, ControlType -> Slider}
]

Is there a better way to do this than the following workaround?

My current workaround is to use a third variable c and Nearest:

DynamicModule[{a = {0, Pi}, b, c},
  {Slider[Dynamic@b, {a}], Dynamic@b, Dynamic[c = First@Nearest[a, b]]}
]

this sets c correctly to 0 or Pi but it is a bit cumbersome.

I cannot reproduce your results with Manipulate in MMA 12.2. The variable does get set to Pi. As an aside, using a slider to choose between only two values is very counterintuitive, since a slider naturally suggests a continuum of values. Use another control that explicitly indicates only two values are available (just like the default Manipulate would do with two values given). — MarcoB, Commented Apr 8, 2021 at 12:42
@MarcoB Ok, then I have to test again with Manipulate. My actual usage includes many multiples of Pi. This question is only a minimal example. — Qbyte, Commented Apr 8, 2021 at 12:48
If you want the control variable to only take exact values you need to specify step values that are exact values, e.g., Manipulate[var, {{var, Pi}, 0, Pi, Pi/16, Appearance -> "Labeled"}] — Bob Hanlon, Commented Apr 8, 2021 at 13:31
@BobHanlon -- You should add this as an answer. Simple and elegant, as we have all come to expect from you;-) — Jagra, Commented Apr 8, 2021 at 15:04

Bob Hanlon · Accepted Answer · 2021-04-08 15:10:03Z

5

If you want the control variable to only take exact values you need to specify all values including the step value as exact values, e.g.,

Manipulate[var, {{var, Pi}, 0, Pi, Pi/16, Appearance -> "Labeled"}]

answered Apr 8, 2021 at 15:10

Bob Hanlon

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$\begingroup$ Thank you for this great answer. However my full problem is that I have a list of numbers which may be also {0, Pi, 4.5, 7 Pi} with no repeatable patterns. Do you also have an easier solution for this case? $\endgroup$
– Qbyte
Commented Apr 8, 2021 at 23:31
$\begingroup$ @Qbyte - For a slider you need to specify the lower bound, upper bound, and step size. You have not said how the list {0, Pi, 4.5, 7 Pi} relates to a slider or what the list is supposed to represent. If you are just trying to select from a list, a SetterBar or RadioButtonBar or PopupMenu may be more appropriate. Provide a complete description of what you are trying to accomplish. $\endgroup$
– Bob Hanlon
Commented Apr 8, 2021 at 23:58

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Jagra · Accepted Answer · 2021-04-08 12:52:47Z

2

Manipulate[If[b < Pi, b, Pi], {b, {0, Pi}, ControlType -> Slider}]

answered Apr 8, 2021 at 12:52

Jagra

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