2016 Update
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Tina Müller
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7965 changed files with 139854 additions and 31002 deletions
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Demonstrate a for-loop where the step-value is greater than one.
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<br><br>
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* Loops/For with a specified step 12/08/2015
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LOOPFORS CSECT
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USING LOOPFORS,R12
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LR R12,R15
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* == Algol style ================ test at the beginning
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LA R3,BUF idx=0
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LA R5,0 from 5 (from-step=0)
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LA R6,5 step 5
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LA R7,25 to 25
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LOOPI BXH R5,R6,ELOOPI for i=5 to 25 step 5
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XDECO R5,XDEC edit i
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MVC 0(4,R3),XDEC+8 output i
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LA R3,4(R3) idx=idx+4
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B LOOPI next i
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ELOOPI XPRNT BUF,80 print buffer
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BR R14
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BUF DC CL80' ' buffer
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XDEC DS CL12 temp for edit
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YREGS
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END LOOPFORS
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* == Fortran style ============== test at the end
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LA R3,BUF idx=0
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LA R5,5 from 5
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LA R6,5 step 5
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LA R7,25 to 25
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LOOPJ XDECO R5,XDEC for j=5 to 25 step 5; edit j
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MVC 0(4,R3),XDEC+8 output j
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LA R3,4(R3) idx=idx+4
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BXLE R5,R6,LOOPJ next j
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XPRNT BUF,80 print buffer
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* == Algol style ================ test at the beginning
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LA R3,BUF idx=0
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LA R5,5 from 5
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LA R6,5 step 5
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LA R7,25 to 25
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DO WHILE=(CR,R5,LE,R7) for i=5 to 25 step 5
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XDECO R5,XDEC edit i
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MVC 0(4,R3),XDEC+8 output i
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LA R3,4(R3) idx=idx+4
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AR R5,R6 i=i+step
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ENDDO , next i
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XPRNT BUF,80 print buffer
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@ -0,0 +1,8 @@
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* == Fortran style ============== test at the end
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LA R3,BUF idx=0
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DO FROM=(R5,5),TO=(R7,25),BY=(R6,5) for i=5 to 25 step 5
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XDECO R5,XDEC edit i
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MVC 0(4,R3),XDEC+8 output i
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LA R3,4(R3) idx=idx+4
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ENDDO , next i
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XPRNT BUF,80 print buffer
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@ -1,20 +0,0 @@
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* Loops/For with a specified step 12/08/2015
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LOOPFORS CSECT
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USING LOOPFORS,R12
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LR R12,R15
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BEGIN LA R3,MVC
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SR R5,R5 index
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LA R6,5 step 5
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LA R7,25 to 25
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LOOPI BXH R5,R6,ELOOPI for i=5 to 25 step 5
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XDECO R5,XDEC
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MVC 0(4,R3),XDEC+8
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LA R3,4(R3)
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NEXTI B LOOPI next i
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ELOOPI XPRNT MVC,80
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XR R15,R15
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BR R14
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MVC DC CL80' '
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XDEC DS CL12
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YREGS
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END LOOPFORS
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FOR i:=5 UNTIL 25 STEP 5 DO
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OUTINTEGER(i)
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for i = 2 to 8 step 2
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print i; ", ";
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next i
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print "who do we appreciate?"
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FOR I = 2 TO 8 STEP 2 : PRINT I; ", "; : NEXT I : PRINT "WHO DO WE APPRECIATE?"
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FOR I = 2 TO 8 STEP 2 : PRINT I; ", "; : NEXT I : PRINT "WHO DO WE APPRECIATE?"
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for i = 2 to 8 step 2
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print i; ", ";
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next i
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print "who do we appreciate?"
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#import system.
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#import extensions.
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#symbol program =
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[
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2 to:8 &by:2 &doEach:i
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[
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console writeLine:i.
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].
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].
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for(i,2,8,2,
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write(i,", ")
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)
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write("who do we appreciate?")
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cat(paste(c(seq(2, 8, by=2), "who do we appreciate?\n"), collapse=", "))
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(do ((i 2 (+ i 2))) ; list of variables, initials and steps -- you can iterate over several at once
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((>= i 9)) ; exit condition
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(display i) ; body
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(newline))
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(let loop ((i 2)) ; function name, parameters and starting values
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(cond ((< i 9)
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(display i)
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(newline)
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(loop (+ i 2)))))) ; tail-recursive call, won't create a new stack frame
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(define-syntax for-loop
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(syntax-rules ()
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((for-loop index start end step body ...)
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(let ((evaluated-end end) (evaluated-step step))
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(let loop ((i start))
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(if (< i evaluated-end)
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((lambda (index) body ... (loop (+ i evaluated-step))) i)))))))
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(for-loop i 2 9 2
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(display i)
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(newline))
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BEGIN
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INTEGER i;
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FOR i:=5 UNTIL 25 STEP 5 DO OutInt(i,5)
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END
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for (( x=2; $x<=8; x=$x+2 )); do
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printf "%d, " $x
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x=2
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while [[$x -le 8]]; do
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echo $x
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((x=x+2))
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done
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for x in {2..8..2}
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do
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echo $x
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x=2
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while ((x<=8)); do
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echo $x
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((x+=2))
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done
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@ -1,3 +1,3 @@
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foreach x (`jot - 2 8 2`)
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echo $x
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end
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for (( x=2; $x<=8; x=$x+2 )); do
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printf "%d, " $x
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done
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for x in {2..8..2}
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do
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echo $x
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done
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foreach x (`jot - 2 8 2`)
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echo $x
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end
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Sub MyLoop()
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For i = 2 To 8 Step 2
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Debug.Print i;
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Next i
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Debug.Print
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End Sub
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buffer=""
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For i = 2 To 8 Step 2
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buffer=buffer & i & " "
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Next
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wscript.echo buffer
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Public Class FormPG
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Private Sub FormPG_Load(sender As Object, e As EventArgs) Handles MyBase.Load
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Dim i As Integer, buffer As String
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buffer = ""
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For i = 2 To 8 Step 2
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buffer = buffer & i & " "
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Next i
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Debug.Print(buffer)
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End Sub
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End Class
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Sub MyLoop()
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For i = 2 To 8 Step 2
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Debug.Print i;
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Next i
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Debug.Print
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End Sub
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