Question

Let's assume that we will run the following Python code:

# initialize myvar - described in each option
print(myvar) # some code - described in each option
print(myvar)

Which of the following options will result in the output of the two print statements being different? Select all that apply.
Note: each choice below first indicates the initialization of myvar. Afterwards, the word Then" is used to separate the #initialize myvar from the #some code
Group of answer choices

A. Assigning a nested list to myvar (type List[List[int]]), then assigning myvar to othervar, and finally assigning a new list at othervar[0].

B. Assigning a list to myvar (type List[str]), then assigning myvar to othervar, and finally assigning a new str at othervar[0].

C. Assigning a list to myvar (type List[str]), then assigning the result of a copy() method on myvar to othervar, and finally calling the replace method on othervar[0].

D. Assigning a list to myvar (type List[str]), then assigning the result of slicing myvar (i.e. myvar[:]) to othervar, and finally calling the replace method on othervar[0].

E. Assigning a list to myvar (type List[str]), then assigning myvar to othervar, and finally using "othervar += ... to add a new element to the list referenced by othervar.

F. Assigning a list to myvar (type List[str]), then assigning the result of slicing myvar (i.e. myvar[:]) to othervar, and finally assigning a new str at othervar[0].

G. Assigning a list to myvar (type List[str]), then assigning the result of a copy() method on myvar to othervar, and finally using "othervar += ... to add a new element to the list referenced by othervar.

H. Assigning a list to myvar (type List[str]), then assigning the result of slicing myvar (i.e. myvar[:]) to othervar, and finally using "othervar += ... to add a new element to the list referenced by othervar.

Answer 1

The options that will result in the output of the two print statements being different are:

C. Assigning a list to myvar (type List[str]), then assigning the result of a copy() method on myvar to othervar, and finally calling the replace method on othervar[0].

E. Assigning a list to myvar (type List[str]), then assigning myvar to othervar, and finally using "othervar += ... to add a new element to the list referenced by othervar.

G. Assigning a list to myvar (type List[str]), then assigning the result of a copy() method on myvar to othervar, and finally using "othervar += ... to add a new element to the list referenced by othervar.

In these cases, the modifications made to the variables will affect their content differently, leading to distinct outputs for the two print statements.

Answer 2

Options A, B, and G will result in different output for the two print statements.

Step-by-step explanation:

The options that will result in the output of the two print statements being different are:

1. Option A: Assigning a nested list to myvar, then assigning myvar to othervar, and finally assigning a new list at othervar[0].
2. Option B: Assigning a list to myvar, then assigning myvar to othervar, and finally assigning a new string at othervar[0].
3. Option G: Assigning a list to myvar, then assigning the result of the copy() method on myvar to othervar, and finally using "othervar += ..." to add a new element to the list referenced by othervar.