line 6 repeated many times in a debugger

    def rec_exponentiation(b,e):
        if e == 0:
            return 1
        else:
            returnrec=rec_exponentiation(b,e-1)
            return b*returnrec
    # Test!
    rec30=rec_exponentiation(3,0)
    rec33=rec_exponentiation(3,3)
    print (rec30)
    print (rec33)

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using a debugger I understand everything
but
when this is called
rec33=rec_exponentiation(3,3)
repeats line6 many times


is this an array

are multiple values returned

in variables i see
returnrec=1
returnrec=3
returnrec=9
and then code goes out of rec_exponentiation function (possibly with value 27)
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rgb192Asked:
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Robert SchuttSoftware EngineerCommented:
That seems to have little to do with Python specifically. You're looking at example code to demonstrate recursion. For this example (calculating 'b' to the power of 'e') recursion is not really necessary but just meant to show you how it's done.

In general the hardest part to determine when using recursion is when to end it. In this case you can see in the code that the recursion ends when e reaches 0. Before that, each time the function calls itself with an 'e' argument 1 less than the one it was called with.

So if you fill in 'b' and 'e' and look at what gets called you would get something like this for the first level:
call    line    code
level   number

0       9       rec33=rec_exponentiation(3,3)

1       2           if 3 == 0:
1       3               return 1
1       4           else:
1       5               returnrec=rec_exponentiation(3,2)
1       6               return 3*returnrec

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Now before returning you can see the function calls itself. So to make this clear, I'll copy the function code, indent it and fill in the new value of 'e'. Do that again until 'e' reaches 0 and you get:
call    line    code
level   number

0       9       rec33=rec_exponentiation(3,3)

1       2           if 3 == 0:
1       3               return 1
1       4           else:
1       5               returnrec=rec_exponentiation(3,2)

2       2                   if 2 == 0:
2       3                       return 1
2       4                   else:
2       5                       returnrec=rec_exponentiation(3,1)

3       2                           if 1 == 0:
3       3                               return 1
3       4                           else:
3       5                               returnrec=rec_exponentiation(3,0)

4       2                                   if 0 == 0:
4       3                                       return 1

3       6                               return 3*returnrec

2       6                       return 3*returnrec

1       6               return 3*returnrec

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Now, when you look at the end there, you will see the reason why the debugger shows line 6 many times (in fact exactly 'e' times).

Note that this is also a reason to be careful with recursion in many languages, each time the function is called it uses stack space (for storing things like function argument and return values) and that space is usually pretty limited.
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rgb192Author Commented:
return 3*returnrec
 return 3*returnrec
 return 3*returnrec
 return 3*returnrec

there is no increment and going to if statement

I do not understand how e is changed because the debugger does not visit other lines
is e an array?
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Robert SchuttSoftware EngineerCommented:
No, e is just a simple (scalar) variable (the second function argument) which on the call rec_exponentiation(3,3) gets initialised with 3, then on the recursive calls is 1 less each time, so 3... 2... 1... 0, and the recursion ends. The return value gets used to return each temporary value, getting larger each time so 1... 3... 9... 27.

Not sure how the python debugger works but usually a debugger only stops on lines with a breakpoint on it. The other lines are being executed though. Maybe stopping on a return statement is a default option in the python IDE if that's what you're using? Can you post a screenshot of the relevant part of the debugger screen? In most debuggers a graphical hint will be displayed in the left margin to show a breakpoint.
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rgb192Author Commented:
debugger stops on all lines

debugger stops on all lines
shown is another file in pyscripter

notice how
a
b
are defined on the left



my question is: I can not follow the debugger for the original .py because it repeats line 6  4 times without going to any other line.

how is value of line6 variable first 1,3,9,27 without changeing by gathering data and incrementing from other lines.
Only example I can think of is an array/list/tuple where 1,3,9,27  are stored
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Robert SchuttSoftware EngineerCommented:
No, it's not an array. The debugging behavior has got to be something specific to the editor/tool you're using. When I run your program (in IDLE) it just writes out 1 and 27 (3 to the power of 0 and 3 to the power of 3).

What should give you interesting information is the tab "Call Stack" (where now Variables is selected in the lower left corner), there you will probably see all the (recursive) calling of functions and parameters.

The reason the variables keep changing is the processing that you suspect is not going on because the debugger isn't stopping but in fact it is still being executed, I can't imagine another reason than breakpoints but that could be due to my unfamiliarity with your editor/debugger. I'll see if I can download a trial version to investigate.
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Robert SchuttSoftware EngineerCommented:
Ok, I see it's a free Google tool. I downloaded it and pasted your code. When I step through with F7, I see all the lines being executed and at some point (see screen capture) I'm at the deepest level and you can see the call stack displaying the function rec_exponentiation having been called 3 times recursively. After that, line 6 is indeed being executed 3 times after another (ending all 3 recursive calls by returning the result so far) but that is exactly what I would expect.
screen capture
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rgb192Author Commented:
call stack:

I think the 2 line 5's should be popped off the stack when debugger reaches line 6 in the function

Maybe that is why I am confused.

I understand why line 9 (calling of the function) stays on the stack but do not understand line 5 (previous line of function and why 2x)
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Robert SchuttSoftware EngineerCommented:
Would it help to add some more textual debugging to the code?
def rec_exponentiation(b,e):
    print 'start of called function rec_exponentiation with b = {}, e = {}'.format(b, e)
    if e == 0:
        print 'e = 0 so we\'re returning 1'
        return 1
    else:
        print 'e = {} so we\'re going deeper'.format(e)
        returnrec=rec_exponentiation(b,e-1)
        print 'the temporary variable returnrec = {} so we will be returning: {}'.format(returnrec, b * returnrec)
        return b*returnrec
# Test!
print 'now calling function: rec_exponentiation(3,0)'
rec30=rec_exponentiation(3,0)
print 'rec30 = ' + str(rec30)
print 'now calling function: rec_exponentiation(3,3)'
rec33=rec_exponentiation(3,3)
print 'rec33 = ' + str(rec33)

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The output shows how the calls and returns follow each other and build up the end result.
The tab "Python interpreter" contains the output:
*** Python 2.7.2 (default, Jun 12 2011, 15:08:59) [MSC v.1500 32 bit (Intel)] on win32. ***
>>> 
*** Remote Interpreter Reinitialized  ***
>>> 
now calling function: rec_exponentiation(3,0)
start of called function rec_exponentiation with b = 3, e = 0
e = 0 so we're returning 1
rec30 = 1
now calling function: rec_exponentiation(3,3)
start of called function rec_exponentiation with b = 3, e = 3
e = 3 so we're going deeper
start of called function rec_exponentiation with b = 3, e = 2
e = 2 so we're going deeper
start of called function rec_exponentiation with b = 3, e = 1
e = 1 so we're going deeper
start of called function rec_exponentiation with b = 3, e = 0
e = 0 so we're returning 1
the temporary variable returnrec = 1 so we will be returning: 3
the temporary variable returnrec = 3 so we will be returning: 9
the temporary variable returnrec = 9 so we will be returning: 27
rec33 = 27
>>> 

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Robert SchuttSoftware EngineerCommented:
I understand why line 9 (calling of the function) stays on the stack but do not understand line 5 (previous line of function and why 2x)
In essence there's no difference between the calling of the function from the main code (adding one level to the stack) and the calling of the same function from within the function itself (also adding to the stack). That's exactly what I tried to visualize in my initial post. At level 3 line 5, you can see that for the last time the function calls itself, then it ends the recursion (because e == 0) and each instance of the function call goes from line 5 to line 6, returning the powers of 3 in increasing order to the function that called it, with the last level returning it into the variable to which the main code is assigning it.
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rgb192Author Commented:
the temporary variable returnrec = 1 so we will be returning: 3
the temporary variable returnrec = 3 so we will be returning: 9
the temporary variable returnrec = 9 so we will be returning: 27


okay emptying out the stack
lifo
this is the only reason why one variable has 3 returns
and one call to function gets the third result
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Robert SchuttSoftware EngineerCommented:
Yep, that's correct.
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rgb192Author Commented:
thanks for stack information.
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