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loop() not implemented? #16

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skilfullycurled opened this issue May 22, 2013 · 8 comments
Closed

loop() not implemented? #16

skilfullycurled opened this issue May 22, 2013 · 8 comments
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@skilfullycurled
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I was creating a simple on/off switch for a sketch using keyPressed() as I often do when debugging and println-ing lots of stuff. I was able to get the sketch to stop using noLoop(), however, when I added loop() the sketch won't even load. When I comment out Loop() the sketch works again.

onOff = true
i = 0;

setup: ->
    size 200, 200

draw: ->
    println i
    i++

keyPressed: ->
    if onOff
        println "turning off"
        noLoop()
    if not onOff
        println "turning on"
        loop()
    onOff = !onOff

I do not believe it is my code because the equivalent JS version works as expected. The following error prints in the IDE console. Firebug does not report anything.

<from source>
org.mozilla.javascript.JavaScriptException: Error: Parse error on line 20: Unexpected ')'
    at org.mozilla.javascript.gen.c246._c0(<CoffeeScriptModeProcessing>:0)
    at org.mozilla.javascript.gen.c246.call(<CoffeeScriptModeProcessing>)
    at org.mozilla.javascript.ContextFactory.doTopCall(ContextFactory.java:398)
    at org.mozilla.javascript.ScriptRuntime.doTopCall(ScriptRuntime.java:3065)
    at org.mozilla.javascript.gen.c246.call(<CoffeeScriptModeProcessing>)
    at org.mozilla.javascript.gen.c246.exec(<CoffeeScriptModeProcessing>)
    at org.mozilla.javascript.Context.evaluateString(Context.java:1104)
    at de.bezier.mode.coffeescript.compiler.CSCompiler.run(Unknown Source)
    at de.bezier.mode.coffeescript.compiler.CSCompiler.compile(Unknown Source)
    at de.bezier.mode.coffeescript.CoffeeScriptBuild.build(Unknown Source)
    at de.bezier.mode.coffeescript.CoffeeScriptBuild.export(Unknown Source)
    at de.bezier.mode.coffeescript.CoffeeScriptMode.handleExport(Unknown Source)
    at de.bezier.mode.coffeescript.CoffeeScriptEditor.handleExport(Unknown Source)
    at de.bezier.mode.coffeescript.CoffeeScriptEditor.serverStarted(Unknown Source)
    at de.bezier.mode.javascript.BasicServer.run(Unknown Source)
    at java.lang.Thread.run(Thread.java:680)
@fjenett
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fjenett commented May 22, 2013

Yeah, i have that on my list ... not sure what's up.

@jashkenas
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Unfortunately. loop is a keyword in CoffeeScript. It loops the block of code until you break out of it.

loop
  print '.'
  break if finished

... so you'll have to figure something out.

@fjenett
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fjenett commented May 24, 2013

Aha, i speculated about that as being the problem but could not find anything in the CS docs right away. Is there a structured documentation that let's me see these things easily? CoffeeScript.org does not list the keywords for example ..

Thanks!

@jashkenas
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Just this sentence:

For readability, the "until" keyword is equivalent to while not, and the "loop" keyword is equivalent to while true.

And the actual list of keywords in the Lexer: http://coffeescript.org/documentation/docs/lexer.html#section-60

@fjenett
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fjenett commented May 24, 2013

K, thanks. I'll fix this later today.

@skilfullycurled
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Not a permanent solution per se but I had started a thread on the Processing forums to confirm that there was a problem with loop() before posting an issue. After incorporating the feedback received here, a user aptly named "GoToLoop" stumbled upon a work around which is to use CS's ability to insert raw JS through backticks. Sure enough, using

`loop()`

will cause loop() to work as expected:

http://forum.processing.org/topic/loop-not-implemented-in-coffeescript-mode#25080000002152759

By the way, does CS mode allow one to access PJS's internal variables?

@fjenett
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fjenett commented May 24, 2013

I added "doLoop" as replacement for "loop" and added a check/warning to the compiler. Plus, there is an example showing the use of both the loop control structure and the noLoop() / doLoop() combo.

Should be closed by commit aca5315 please update your mode using the mode manager.

@fjenett fjenett closed this as completed May 24, 2013
@skilfullycurled
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sweet! thank you my friend.

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@jashkenas @fjenett @skilfullycurled