Hamido Hamido - 1 year ago 54
Python Question

Python recursive function call with if statement

I have a question regarding function-calls using if-statements and recursion.
I am a bit confused because python seems to jump into the if statements block even if my function returns "False"

Here is an example:

1 def function_1(#param):
2 if function_2(#param):
3 #do something
4 if x<y:
5 function_1(#different parameters)
6 if x>y:
7 function_1(#different parameters)

My function_2 returns "False" but python continues the code on line 5 for example. Can anyone explain this behavior? Thanks in advance for any answers.

edit: Sorry, forgot the brackets

concrete example:

1 def findExit(field, x, y, step):
2 if(isFieldFree(field, x, y)):
3 field[y][x] = filledMarker
4 findExit(field, x + 1, y, step+1)
5 findExit(field, x - 1, y, step+1)
6 findExit(field, x, y + 1, step+1)
7 findExit(field, x, y - 1, step+1)
8 elif(isFieldEscape(field, x, y)):
9 way.append(copy.deepcopy(field))
10 wayStep.append(step+1)

def isFieldFree(field, x, y):
if field[y][x] == emptyMarker:
return True
return False

def isFieldEscape(field, x, y):
if field[y][x] == escapeMarker:
return True
return False

After both functions "isFieldFree" and "isFieldEscape" return False python continues the code in line 5 sometimes in line 6.

Answer Source

You may misunderstand how recursion works, yes it continues at line 5 or 6 because the recursion has ended at a lower level in the call stack, so it continues at a higher-level in the call stack. Here's a sample call stack, note the next operation after False is the next findExit() at the higher call stack:

1 findExit(...):
2    True:
3        field assignment
4.1      findExit(x+1) 
  2          True
  3              field assignment
  4.1            findExit(x+1):
    2                False  # Does not jump to line 5 in current call stack.
  5.1            findExit(x-1):
    .                ...
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