Forum Discussion

Komag's avatar
Komag
Roku Guru
6 years ago

How to skip an iteration in a FOR loop?

FUNCTION testNext()
	testA = ["apple", 2, "carrot", "dapple", 5, "fapple", "gourd"]
	testInt = 0
	FOR EACH fruit IN testA
		testInt++
		IF TYPE(fruit) = "roString" THEN ? fruit ELSE testA.Next()
		? "testNext() did one step, fruit " fruit ", testInt" testInt
	END FOR
END FUNCTION

This doesn't do what I want, it increments the counter but doesn't skip the rest of the loop cycle, and then the counter is incremented AGAIN at the bottom, so "carrot" never prints out, etc. I'd like to avoid using GOTO if possible. Any ideas?
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  • destruk's avatar
    destruk
    Streaming Star
    6 years ago
    FUNCTION testNext()
       testA = ["apple", 2, "carrot", "dapple", 5, "fapple", "gourd"]
       testInt = 0
       FOR EACH fruit IN testA
          testInt++
          IF TYPE(fruit) = "roString" THEN
               ? fruit
          ELSE
               Next
          END IF
          ? "testNext() did one step, fruit " fruit ", testInt" testInt
       END FOR
    END FUNCTION
  • speechles's avatar
    speechles
    Roku Guru
    6 years ago
    What exactly are you trying to do? Why are you increment in the for each loop?

    FUNCTION testNext()
       testA = ["apple", 2, "carrot", "dapple", 5, "fapple", "gourd"]
       for fruit = 0 to testA.count()-1
           if type(testA[fruit]) = "roString" ? testA[fruit]
           ? "testNext(): did one step; fruit="+testA[fruit].toStr()+" index="+fruit.toStr()
       next fruit
    END Function

    Wouldn't that be easier.. I mean sorta? Are you just trying to build a string of names of fruit from among a non sorted one dimensional array? What exactly are you trying to do sir.. lol

    Is shorter better or does adding that extra reference to the array mean it slows down the code and longer is better. You be the judge. you need a huge array to test the millisecond differences. In a list small as this the difference is NIL.

    ----------------------------------------------------------------
    For some reason the Roku docs do not note the special use of "next varname" as an option.

    Take this for example:
    for fruit = 0 to testA.count()-1
    for potion = 0 to testB.count()-1
    for skill = 0 to testC.count()-1

    Now at any time we can "Next fruit" or "Next potion" or "Next skill" and not involve the other nested for loops incrementing. Using "End for" will interact with the last nested for. Using "next varname" will increment the loop for the associated varname.

    Why isn't this in the Roku documentation?

    RokuTannerD This is an oversight and should be corrected. You can use "Next varname" when using "For varname=" in a loop. When doing this it allows incrementing nested for's out of order which the "End for" does not allow. Thanks. ^_~

    https://developer.roku.com/docs/references/brightscript/language/program-statements.md

    it is that page that is incorrect --^

    Also.. why can't we

    For each {varname1 varname2 varname3} in var ? Other languages allow this but not brightscript.
  • Komag's avatar
    Komag
    Roku Guru
    6 years ago
    destruk, that gives me:

    Syntax Error. (compile error &h02) in pkg:/source/mn.brs(1846)

    which is the line with Next
  • Komag's avatar
    Komag
    Roku Guru
    6 years ago
    speechles, that's very interesting about using NEXT Var in place of END FOR.

    (The integer increment was just for information on the printout to help me clarify what was going on.)

    Basically I'm trying to process objects in an array, and if the object has a certain property, do all the rest of the processing, otherwise I want to skip to the next object. My current solution is to simply make it a huge IF block, indenting over dozens of lines of code, which aesthetically I'd rather not do but it works just fine. So really this thread is more academic and not really important.
  • destruk's avatar
    destruk
    Streaming Star
    6 years ago
    "Komag" wrote:
    destruk, that gives me:

    Syntax Error. (compile error &h02) in pkg:/source/mn.brs(1846)

    which is the line with Next

    That is odd.  Perhaps they can not be mixed, but all of my loops I write for my own apps use Next instead of End For.
    So I'll need to test and see if replacing End For here with Next resolves that.
    Next is a reserved word in brightscript and I was always under the impression they (End For and Next) are interchangeable.

    FUNCTION testNext()
       testA = ["apple", 2, "carrot", "dapple", 5, "fapple", "gourd"]
       testInt = 0
       FOR EACH fruit IN testA
          testInt++
          IF TYPE(fruit) = "roString" THEN
               ? fruit
          ELSE
               Next
          END IF
          ? "testNext() did one step, fruit " fruit ", testInt" testInt
       Next
    END FUNCTION
  • destruk's avatar
    destruk
    Streaming Star
    6 years ago
    "squirreltown" wrote:
    NEXT does not work for me  inside that if statement. Syntax error.

    Weird, ok, you are right.  It can't be within the IF statement.
    But this works fine if you rewrite the loop slightly.  And if you're wanting the index value of the array for the printed results, you should increment the value after the print statement.
    Sub Main()
       testA = ["apple", 2, "carrot", "dapple", 5, "fapple", "gourd"]
       testInt = 0
       FOR EACH fruit IN testA
          IF TYPE(fruit) = "roString" THEN
               ? fruit
    	   ? "testNext() did one step, fruit " fruit ", testInt" testInt
          END IF
          testInt++
      Next
    End Sub
  • destruk's avatar
    destruk
    Streaming Star
    6 years ago
    Just for kicks I tried, and yes, End For can't be within the If statement either, so it's consistent. Other languages (like PHP) allow this - next will exit any subsequent nested if statements and increment the internal loop counter and re-execute the inner-most loop, just not brightscript.
  • belltown's avatar
    belltown
    Roku Guru
    6 years ago
    Why not just let the loop construct handle the counter:


    Sub Main()
       testA = ["apple", 2, "carrot", "dapple", 5, "fapple", "gourd"]
       FOR i = 0 TO testA.Count() - 1
          fruit = testA[i]
          IF TYPE(fruit) = "roString" THEN
               ? i; ": "; fruit
          END IF
        END FOR
    End Sub



     0: apple
     2: carrot
     3: dapple
     5: fapple
     6: gourd
  • Komag's avatar
    Komag
    Roku Guru
    6 years ago
    My whole point was to NOT process stuff sometimes, so that the counter results would be sequential, such as:
     1: apple
     2: carrot
     3: dapple
     4: fapple
     5: gourd

    but apparently that can only be done by working with big IF statements and lots of indenting!
    FUNCTION testNext()
       testA = ["apple", 2, "carrot", "dapple", 5, "fapple", "gourd"]
       testInt = 0
       FOR EACH fruit IN testA
          IF TYPE(fruit) = "roString"
             testInt++
             ? "testNext() found a String, fruit " fruit ", testInt" testInt
             ' And do 50 more lines of code,
             ' all indented to be within this huge IF block,
             ' instead of just skipping to next item in list,
             ' too bad
             ' Could maybe use GOTO but that's not something I like to do,
             ' So the IF block with indenting is the lesser of two evils!
          END IF
       END FOR
    END FUNCTION