HOTSPOT -
You create the following program to locate a conference room and display the room name. Line numbers are included for reference only.
Colleagues report that the program sometimes produces incorrect results.
You need to troubleshoot the program. Use the drop-down menus to select the answer choice that answers each question based on the information presented in the code segment.
Hot Area:
References:
https://www.w3resource.com/python/python-data-type.php
https://www.w3resource.com/python/python-if-else-statements.php
3rd one should be mismatch data type..
Is not invalid syntax. The syntax is fine. The line if not room in rooms: does not throw any errors. The problem is that room is a string and the dictionary rooms keys are integers so a match will never be found. The problem is a mismatch data type.
It is asking to enter Room number, not room name. so we give 1 or 2 which matches with Dict rite?
In line 2, room is a string. So even if you enter 1 or 2, it will be considered a string. Since line 2 input only takes in string values. So in line 3, it is comparing string 2 to int 2, which will cause a mismatch data type error
In line 2, room is a string. So even if you enter 1 or 2, it will be considered a string. Since line 2 input only takes in string values. So in line 3, it is comparing string 2 to int 2, which will cause a mismatch data type error
100% confident that: int and string string mismatched data type(s)
It won't find the room number because the input returns a string value that it's using to match against the int keys in the dictionary (e.g. "1" can't find the first room with a key of 1). So the issue is a mismatched data type. However, I also see a second issue here, in the else statement it uses the room variable to print the room name from rooms, but the room numbers are 1 and 2, whereas the indexes for those rooms are 0 and 1.
There isn't a second issue because it is python dictionaries. so it will see 1 and 2.
Line 03 - if not room in rooms: The problem is with the above line it is syntax issue - as the program won't compile, it should be "if room not in rooms:"
Invalid Syntax is correct. If we want to iterate on the Keys or items we need to use dictionary.keys() or dictionary.items() to loop on them. In the question it should be "room in rooms.keys()", so its a invalid Syntax.
Nope. If you simply update line 02 to take int as input, the code will display the room name. No issue with line 03 syntax. I have executed the code below and it worked! rooms = {1: 'Lab', 2: 'Conf'} room = int(input('enter the room number: ')) if room not in rooms: print('room number does not exist') else: print("the room name is: " + rooms[room])
Nope. If you write the code like this rooms = {'1': 'Foyer', '2': 'Conference'} then you will be getting the answer like "The room name is Foyer". Anyway rooms.keys() returns the datatype is int. but room variable's datatype is string. So answer should be mismatched data type(s) though i am not getting any error only getting "room number does not exist"
3rd one should actually be Mismatch Data Type since the code doesn't actually give any errors for invalid syntax. It works fine but the real problem is that the user input will always be in "string" form no matter how they type it in. It needs to be converted into "integer" form.
rooms = {1: 'Foyer , 2: 'Conference room'} ^ SyntaxError: invalid syntax
You missed to add ' after Foyer, that's why. The answer should be mismatch data your
Line 03 fails because of both syntax error as well as data type mismatch: 1. line 02 needs integer conversion of the string: room=int(input('Enter the room number: ')) 2.Line 03 needs syntax: if room not in rooms: ------------------------ Not sure which one to choose? P.S.ITs a dictionary,so rooms[room] is referring to the key value unlike a list which has fixed index numbering that too starting at 0
first program will try to check syntax(if room not in rooms:) then data type so if we need to choose 1 option it will be invalid syntax
correct answer
All correct
How is it marked in the test?