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Results tagged with Search options user 280
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Questions about how Mathematica evaluates expressions, tracing evaluation (Trace*), handling expressions in unevaluated form (Hold*, Unevaluated), nonstandard evaluation, etc.

Starting from Mathematica version 6 Show by default has no side effects. In your particular case it just returns the Graphics expression to StepMonitor which does not use it because the latter is desi …
answered Sep 26 '15 by Alexey Popkov
How exactly does Mathematica determine that evaluation of particular expression should be finished and the result should be returned? Here are some examples of unclear behavior which have arisen … , when I tried to understand deeper Todd Gayley's Block trick: x := Block[{tried = True}, x + 1] /; ! TrueQ[tried] x + y (* 1 + x + y *) x + 1 During evaluation of In:= \$IterationLimit …
asked Mar 17 '11 by Alexey Popkov
WReach has presented here a nice way to represent the Mathematica's evaluation sequence using OpenerView. It is much more clear way to go than using the standard Trace or TracePrint commands. But it … evaluation subsequence begins and from which expression (it is better to have each subsequence exactly in one Opener). The evaluation (sub)sequence should be identified as easily as possible with the …
asked Mar 28 '11 by Alexey Popkov
It is because a[] does not evaluate to v1 in the case of assignment a[] = 0 because Set (=) has attribute HoldFirst: In:= Set // Attributes Out= {HoldFirst, Protected, SequenceHold} So …
answered Jan 10 '13 by Alexey Popkov
When trying to make an optimized version of my plotRange function I faced a problem of catching the "Preemptive" evaluations generated by the FrontEnd from withing the main evaluation loop. Consider … ]; {xRange, yRange} NotebookClose[nb] {xRange, yRange} {{-0.392699, 19.2423}, {-1.04167, 1.04167}} One can see that the first evaluation of {xRange, yRange} returns undefined symbols because …
asked Jan 23 '13 by Alexey Popkov
You could use InString for that: ToExpression[InString, StandardForm, Hold] Hold[xyPoints = {{1, 2}, {2, 5}, {3, 10}}; ListLinePlot[xyPoints]]
answered Aug 14 '13 by Alexey Popkov
I try with Ctrl+A, then Ctrl+click to deselect unnecessary cells and then Enter but mathematica evaluate only one cell. You should use the Evaluation ▶ Evaluate Cells menu item: http …
answered Oct 8 '17 by Alexey Popkov
Try the Evaluated -> False option: FindRoot[Print[x]; Sin[x] - Cos[x], {x, .5}, Evaluated -> False] During evaluation of In:= 0.5 During evaluation of In:= 0.5 During evaluation … of In:= 0.5 During evaluation of In:= 0.793408 During evaluation of In:= 0.793408 During evaluation of In:= 0.793408 During evaluation of In:= 0.785398 During …
answered Jun 9 '12 by Alexey Popkov
Set is specially overloaded for Part in order to allow in-place modification of the expressions: a = x -> 2; a[] = 1; a 1 -> 2 It is documented on the Documentation page for Part (the first …
answered May 16 '17 by Alexey Popkov
Another way is to use delayed assignment in the first argument of With: With[{init := {a = 1, b = 2}}, With[init, {a, b}]] {1, 2} Or if you prefer to store your variable specification list as …
answered Aug 22 by Alexey Popkov
What you see in the output cell is just a visualization of the actual result of the evaluation. When working with the FrontEnd the default FormatType of the output stream is StandardForm what means … evaluation by requesting it from the history: In:= HoldComplete[a 2 1 2]; In:= % // FullForm Out//FullForm= HoldComplete[Times[a,2,1,2]] Related: Conversion of expressions by the FrontEnd …
answered May 27 '17 by Alexey Popkov
You can use the Evaluator option of Cell for this. You should define several kernels through the Kernel Configuration Options... dialog, then you can use a name of configured kernel as value for the E …
answered Aug 25 '13 by Alexey Popkov
evaluation of the arguments before applying the rules, and this is exactly what happens in the case of NIntegrate: NIntegrate[x + x, {x, 1, 2}] // Trace {NIntegrate[x + x, {x, 1, 2}], {x + x, 2 x}, {{x} =., {x =.}, {x =., Null}, {Null}}, {x =., Null}, 3.} …
answered Jun 9 '16 by Alexey Popkov
That's because Plot is a numerical function which evaluates its first argument with numerical values of x while FourierSinSeries requires symbolic variable. What happens is that Plot internally makes …
answered Mar 16 '17 by Alexey Popkov
In addition to Mr.Wizard's answer here is a collection of other possibilities: data1 = {1, 1}; data2 = {2, 2}; datalist := {data1, data2}; ToString /@ Map[HoldForm, OwnValues[datalist], {3}][[1, 2]] …
answered Mar 13 '17 by Alexey Popkov

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