The following recursive function is supposed to return a list that is the reverse of the list it was given. But it has a bug What is the bug? (Line numbers are not part of the code.) 1. def revlist(Ist): 2. if len(Ist) == 0: 3. return Ist 4. else: 5. return revlist(Ist[1:]) + [Ist[0]] Line 5 should use Ist.append( instead of concatenation. O Line 4 should be elif:, not else: Line 2 should be: if Ist == []: O The parameter Ist should be list. O The base case should be: if len(Ist) == 0 or len(Ist) == 1:

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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The following recursive function is supposed to return a list that is the reverse of the list it was given. But it has a bug.
What is the bug? (Line numbers are not part of the code.)
1. def revlist(Ist):
2.
if len(Ist) == 0:
3.
return Ist
4.
else:
5.
return revlist(lst[1:]) + [Ist[0]
O Line 5 should use Ist.append() instead of concatenation.
O Line 4 should be elif:, not else:
O Line 2 should be: if Ist == []:
O The parameter Ist should be list.
O The base case should be: if len(Ist)
== 0 or len(Ist) == 1:
Question 7
What is the output of running the following program?
def showme(arg):
if arg < 10:
Transcribed Image Text:The following recursive function is supposed to return a list that is the reverse of the list it was given. But it has a bug. What is the bug? (Line numbers are not part of the code.) 1. def revlist(Ist): 2. if len(Ist) == 0: 3. return Ist 4. else: 5. return revlist(lst[1:]) + [Ist[0] O Line 5 should use Ist.append() instead of concatenation. O Line 4 should be elif:, not else: O Line 2 should be: if Ist == []: O The parameter Ist should be list. O The base case should be: if len(Ist) == 0 or len(Ist) == 1: Question 7 What is the output of running the following program? def showme(arg): if arg < 10:
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