CODEWITHSUNDEEP
python-3.x
user1162657
user1162657

Reputation: 80

Showing that function for IF statement does not work the same way as IF statement

This is a question for a homework problem that I cant figure out:

Question Beginning

Q3. Let's try to write a function that does the same thing as an if statement:

def if_function(condition, true_result, false_result):

    """Return true_result if condition is a true value, and false_result otherwise."""

    if condition:
        return true_result
    else:
        return false_result

This function actually doesn't do the same thing as an if statement in all cases. To prove this fact, write functions c, t, and f such that one of these functions returns the number 1, but the other does not:

def with_if_statement():

    if c():
        return t()
    else:
        return f()

def with_if_function():

    return if_function(c(), t(), f())

Question End

Heres what I figured out:

with_if_statement() does not evaluate f() if c() is true, but with_if_function() evaluates all 3 before checking if c() is true or not.

So, I thought of assigning a global variable in c(), and changing its value in f()

heres my code (which does not work):

def c():

    try:
        global x
    except NameError:
        x=1
    if x==1:
        return True
    else:
        return False

def t():

    if x==1:
        return (1)
    else:
        return (0)

def f():

    global x
    x=2
    if x==1:
        return (1)
    else:
        return (0)

can anyone help me figure out the answer? Thanks..!

Upvotes: 4

Views: 6404

Answers (2)

Kulbear
Kulbear

Reputation: 901

def if_function(condition, true_result, false_result):
    """Return true_result if condition is a true value, and
    false_result otherwise.

    >>> if_function(True, 2, 3)
    2
    >>> if_function(False, 2, 3)
    3
    >>> if_function(3==2, 3+2, 3-2)
    1
    >>> if_function(3>2, 3+2, 3-2)
    5
    """
    if condition:
        return true_result
    else:
        return false_result


def with_if_statement():
    """
    >>> with_if_statement()
    1
    """
    if c():
        return t()
    else:
        return f()

def with_if_function():
    return if_function(c(), t(), f())

The question requires that, write 3 functions: c, t and f such that with_if_statement returns 1 and with_if_function does not return 1 (and it can do anything else)

At the beginning, the problem seems to be ridiculous since, logically, with_if_statement and with_if_function are same. However, if we see these two functions from an interpreter view, they are different.

The function with_if_function uses a call expression, which guarantees that all of its operand subexpressions will be evaluated before if_function is applied to the resulting arguments. Therefore, even if c returns False, the function t will be called. By contrast, with_if_statement will never call t if c returns False (This paragraph from the UCB website).

def c():
    return True

def t():
    return 1

def f():
    '1'.sort() # anything breaks the program is OK

Upvotes: 2

Greg Hewgill
Greg Hewgill

Reputation: 993951

The global statement shouldn't throw a NameError (and so you won't run x=1 in c()). I would try rewriting your code without using exceptions, they won't be necessary to solve this and are making it more complicated than it needs to be. Using a global variable and having side effects in your functions is certainly the right track.

Upvotes: 2

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