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Question 25

What is the output of the following code?

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def bar(f, a, b):
    return f(a, b)

a, b = 2, 3
x = bar(lambda a,b:a*b, bar(lambda a,b:a+b, a, b), bar(lambda a,b:a-b, a, b))
print(x)
Hint

Trace the values of f, a, and b inside bar() carefully. f seems to represent a reference to a function, whilst a and b, on first glance, may be integer values.

Solution

Given x = bar(lambda a,b:a*b, bar(lambda a,b:a+b, a, b), bar(lambda a,b:a-b, a, b)),

x is the result of f(a, b) where f = lambda a, b: a * b, a = bar(lambda a,b:a+b, a, b), and b = bar(lambda a,b:a-b, a, b).

Function f is? a b
lambda a, b: a * b bar(lambda a,b:a+b, a, b) bar(lambda a,b:a-b, a, b)

For parameter a = bar(lambda a,b:a+b, a, b), obtain the result of plugging in values of global scope a and b into lambda a, b: a + b \(\Rightarrow\) 2 + 3 = 5.

For parameter b = bar(lambda a,b:a-b, a, b), obtain the result of plugging in values of global scope a and b into lambda a, b: a - b \(\Rightarrow\) 2 - 3 = -1.

We now have:

Function f is? a b
lambda a, b: a * b 5 -1

f(a, b) = (lambda a, b: a * b)(5, -1) = 5 * -1 = -5.

Answer 
-5