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codebreaker exercise visualsteps
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codebreaker exercise visualsteps
#2
I have trouble understanding the question. However, I suggest to type into interactive interpreter:

>>> help('for') 
The "for" statement
*******************

The "for" statement is used to iterate over the elements of a sequence
(such as a string, tuple or list) or other iterable object:

   for_stmt ::= "for" target_list "in" expression_list ":" suite
                ["else" ":" suite]

The expression list is evaluated once; it should yield an iterable
object.  An iterator is created for the result of the
"expression_list".  The suite is then executed once for each item
provided by the iterator, in the order returned by the iterator.  Each
item in turn is assigned to the target list using the standard rules
for assignments (see Assignment statements), and then the suite is
executed.  When the items are exhausted (which is immediately when the
sequence is empty or an iterator raises a "StopIteration" exception),
the suite in the "else" clause, if present, is executed, and the loop
terminates.

A "break" statement executed in the first suite terminates the loop
without executing the "else" clause’s suite.  A "continue" statement
executed in the first suite skips the rest of the suite and continues
with the next item, or with the "else" clause if there is no next
item.

The for-loop makes assignments to the variables(s) in the target list.
This overwrites all previous assignments to those variables including
those made in the suite of the for-loop:

   for i in range(10):
       print(i)
       i = 5             # this will not affect the for-loop
                         # because i will be overwritten with the next
                         # index in the range

Names in the target list are not deleted when the loop is finished,
but if the sequence is empty, they will not have been assigned to at
all by the loop.  Hint: the built-in function "range()" returns an
iterator of integers suitable to emulate the effect of Pascal’s "for i
:= a to b do"; e.g., "list(range(3))" returns the list "[0, 1, 2]".

Note: There is a subtlety when the sequence is being modified by the
  loop (this can only occur for mutable sequences, e.g. lists).  An
  internal counter is used to keep track of which item is used next,
  and this is incremented on each iteration.  When this counter has
  reached the length of the sequence the loop terminates.  This means
  that if the suite deletes the current (or a previous) item from the
  sequence, the next item will be skipped (since it gets the index of
  the current item which has already been treated).  Likewise, if the
  suite inserts an item in the sequence before the current item, the
  current item will be treated again the next time through the loop.
  This can lead to nasty bugs that can be avoided by making a
  temporary copy using a slice of the whole sequence, e.g.,

     for x in a[:]:
         if x < 0: a.remove(x)

Related help topics: break, continue, while
(END)
I'm not 'in'-sane. Indeed, I am so far 'out' of sane that you appear a tiny blip on the distant coast of sanity. Bucky Katt, Get Fuzzy

Da Bishop: There's a dead bishop on the landing. I don't know who keeps bringing them in here. ....but society is to blame.
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Messages In This Thread
codebreaker exercise visualsteps - by ivobrugman - Aug-10-2019, 08:12 PM
RE: codebreaker exercise visualsteps - by perfringo - Aug-11-2019, 08:35 AM
RE: codebreaker exercise visualsteps - by ThomasL - Aug-11-2019, 09:44 AM

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