Having trouble with a while loop, it breaks, even though it shouldn't

Having trouble with a while loop, it breaks, even though it shouldn't

By : user2952941
Date : November 19 2020, 03:59 PM
Hope this helps I'm having trouble with a while loop! , Your validation part can be more simple Try something like this
code :
while x == 1:
    user = raw_input(">> ")
    usernames = [i.split()[0] for i in passwr]
    if user.lower() not in usernames:
            x = 0

            print 'Sorry, that\'s already in use!'
What's your username?
>> Username1
Sorry, that's already in use!
>> Username2    
Sorry, that's already in use!
>> Username3
What's your pw?
>> Password
username1 Password
username2 Password
Username3 Password

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Java applet double buffer trouble (and loop trouble)

Java applet double buffer trouble (and loop trouble)

By : SmugJustin
Date : March 29 2020, 07:55 AM
hop of those help? Put a check for size in the paint method. If the size has changed, create a new image and then draw to it, and store the new size information. Also, don't call repaint as you are just have a recursive call to paint.
If you put an unending loop in the start method then the thread never exits the start method.
code :
public void paint(Graphics g) 

        Dimension currentSize  = getSize();
        if ( ! currentSize.equals(size) ) {
          size = currentSize;
          buf = createImage(size.width,size.height); 
          g2d = buf.getGraphics();
        //g2d = (Graphics2D) g;
        g2d.drawString("Welcome 2 Java!!", (int) 50, 60 );
        //g2d.drawString("Welcome to Java!!", (int) p1.x, 60 );

        g.drawImage(buf, 0, 0, this);

The proper Python exception to break out of an infinite loop (that shouldnt be infinite)

The proper Python exception to break out of an infinite loop (that shouldnt be infinite)

By : Ian D
Date : March 29 2020, 07:55 AM
it should still fix some issue The only way this wouldn't occur is if do_stuff never returns false.
If that is the case create a custom exception. I usually create a MiscException (along with more specific exceptions)
trouble with 'breaks' in for loops in R

trouble with 'breaks' in for loops in R

By : user3261391
Date : March 29 2020, 07:55 AM
I wish this help you Since answer cannot be true if while loop is not broken, put the answer value check after while loop and break for loop if answer is true:
code :
x <- c(3, -7, 6, 3, 1, -7)
n <- length(x); n # = 6
counter <- 0
answer <- FALSE
for (i in 1:(n-1)) {
    v <- i + 1
    while (v <= n) {
      counter <- counter + 1
      if (x[i] == x[v]) {
        answer <- TRUE
      else {
        v <- v + 1

    } #end while

    if(answer == TRUE)         

} #end for
answer # = TRUE
counter # = 13, SHOULD BE 3
Having trouble with continue/breaks in javascript

Having trouble with continue/breaks in javascript

By : Leandro Camargo
Date : March 29 2020, 07:55 AM
wish of those help Your break/continue are inside child functions. break/continue only apply to loops at the SAME code level, but since you're executing them inside sub-functions where there are no loops, there's nothing to break/continue, and they're effectively "do nothing" statements.
in more detail:
How to loop the split of different cells with line breaks into one cell with line breaks (VBA Excel)

How to loop the split of different cells with line breaks into one cell with line breaks (VBA Excel)

By : dragonstark
Date : March 29 2020, 07:55 AM
wish of those help I tested this code on 10 rows and it works as expected but Column E will need to be resized manually. It seems that Columns("E").AutoFit is not working here due to the presence of Chr(10)
code :
Option Explicit

Sub Test()

Dim SplitA, SplitB, SplitC, SplitD
Dim i As Long, j As Long

Dim Final As String

For i = 2 To Range("A" & Rows.Count).End(xlUp).Row
    SplitA = Split(Range("A" & i), Chr(10))
    SplitB = Split(Range("B" & i), Chr(10))
    SplitC = Split(Range("C" & i), Chr(10))
    SplitD = Split(Range("D" & i), Chr(10))

        For j = LBound(SplitA) To UBound(SplitA)
            Final = Final & SplitA(j) & Chr(32) & SplitB(j) & Chr(32) & SplitC(j) & Chr(32) & SplitD(j) & Chr(32) & Chr(10)
        Next j

        Range("E" & i) = Left(Final, Len(Final) - 2)

    SplitA = ""
    SplitB = ""
    SplitC = ""
    SplitD = ""
    Final = ""
Next i

End Sub
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