How are functions called in a Recursive function when there are more than 1 function call in one statement in Python?

Question

def fibonacci(n):

    if n == 0 or n == 1: # base c a s e s
        return 1
    else:
        return fibonacci(n-1) + fibonacci(n-2)

print(fibonacci(10))

When fibonacci(10) is called , is fibonacci(n-1) resolved first till the base case and after that,fibonacci(n-2) is resolved?

You could find out by adding print(‘fibonacci ‘ + str(n)) at the top of the function. — , Jun 27 '20 at 12:16

score 1 · Accepted Answer · answered Jun 27 '20 at 12:21

From the docs:

Python evaluates expressions from left to right.

So yes, the evaluation order for fibonacci(n-2) + fibonacci(n-1) is as you describe: first fibonacci(n-2) is evaluated to a value, then fibonacci(n-1) is evaluated to a value, and the entire expression evaluates to the sum of these two values.

There’s nothing special about recursion in this case. Python evaluates left-to-right even if the operands aren’t recursive function calls.

How are functions called in a Recursive function when there are more than 1 function call in one statement in Python?

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