I am looking at a simple recursive function, and oxymoron aside, I am having
difficulty in seeing what occurs. I have tried adding some debug print commands
to help break the thing down. This helps a lot, but I still have a question
that I need help with.
Here is original code:
def mult(a, b):
if b == 0:
return 0
rest = mult(a, b - 1)
value = a + rest
return value
print "3 * 2 = ", mult(3, 2)
I see how python outputs the string "mult(3,2)" before running the function,
and then adds the return value to the output,
and I see how the 'if' condition is met,
but I do not see how the variable 'rest' is assigned a value, and a single
value at that.
'rest' must have a value of 3 (for this example) for the function to return the
correct
value, and 'rest' must be assigned the value of '3' but when, prior, to b == 0?
When 'b' == 0 (zero) the 'if' condition is escaped.
In the first iteration, 'rest' is shown with the value '0' (zero),
and in the second it is '3'.
1st, run rest = mult(3, 1)
2nd. run rest = mult(3, 0) and the 'if' condition is met
Is there some inherent multiplication e.g. 3*1 = 3 and 3*0 = 0,
and even if so, it seems backwards?
I am once again perplexed!
Help please. . . if you can.
-A
Here is my debugging code:
#! /usr/bin/env python
def mult(a, b):
if b == 0:
return 0
print
print 'a =', a
print 'b =', b
rest = mult(a, b - 1)
print
print 'rest =', rest
value = a + rest
print 'value =', value, '(a + rest ==',a, '+', rest,')'
return value
print
print 'mult(3,2) value = ', '\n', '\nreturn = ' + str(mult(3, 2))
Here is the output from the debugging code:
mult(3,2) value =
a = 3
b = 2
a = 3
b = 1
rest = 0
value = 3 (a + rest == 3 + 0 )
rest = 3
value = 6 (a + rest == 3 + 3 )
return = 6
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