Python recursive function call with if statement

问题

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
        else:
            return False

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

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

回答1:

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):
    .                ...

回答2:

Short answer:

That's because you're not actually calling the function. You can call the function by using the parenthesis.

if function2():
    ...

Long answer:

Functions in Python are a first class citizen (functional paradigm), and so it is completely valid to refer to a function just by its name. The following is valid syntax:

def hello():
    print("Hello")

hello_sayer = hello
hello_sayer() # print "Hello"

The next concept in play is truth-ness of non-Boolean variables. In Python, the following are considered False-y

None
False
zero of any numeric type, for example, 0, 0L, 0.0, 0j.
any empty sequence, for example, '', (), [].
any empty mapping, for example, {}. instances of user-defined classes, if the class defines
a nonzero() or len() method, when that method returns the integer zero or bool value False.

Everything else is True-ish. Since a function name falls under none of the above categories, it is considered True-ish when tested in a conditional context.

Reference: https://docs.python.org/3/library/stdtypes.html#truth-value-testing

Edit: The earlier question was incomplete, and did not have a function call. For the new question, AChampion's answer is the correct one.

来源：https://stackoverflow.com/questions/44871858/python-recursive-function-call-with-if-statement

标签

python

function

if-statement

recursion