Knowee
Questions
Features
Study Tools

How many different objects of type list exist in the computer’s memory after the followingcode has been executed? Assume that no garbage collection has been executed and that no optimizationshave been performed.import copya = [1, 2, 3, 9001]b = []for x in a:b = b + [x]c = aa = b[:]b = copy.deepcopy(b)

Question

How many different objects of type list exist in the computer’s memory after the followingcode has been executed? Assume that no garbage collection has been executed and that no optimizationshave been performed.import copya = [1, 2, 3, 9001]b = []for x in a:b = b + [x]c = aa = b[:]b = copy.deepcopy(b)

🧐 Not the exact question you are looking for?Go ask a question

Solution

After the code has been executed, there are 4 different list objects in the computer's memory. Here's the step-by-step explanation:

  1. a = [1, 2, 3, 9001]: This creates the first list object in memory.

  2. b = []: This creates the second list object in memory.

  3. for x in a: b = b + [x]: This loop creates a new list object in memory for each iteration because the + operation between lists results in a new list. Since a has 4 elements, this loop creates 4 new list objects. However, only the final list object (which is equivalent to a) is referenced by b after the loop ends. The other 3 intermediate list objects are not referenced by any variable.

  4. c = a: This does not create a new list object. It just creates a new reference c to the existing list object referenced by a.

  5. a = b[:]: This creates a shallow copy of b, resulting in the third unique list object in memory.

  6. b = copy.deepcopy(b): This creates a deep copy of b, resulting in the fourth unique list object in memory.

So, in total, there are 4 different list objects in the computer's memory after the code has been executed.

This problem has been solved

Similar Questions

Consider the following code. Assume that copy has been imported.w = [1, 2]z = copy.copy(w)z = copy.deepcopy(w)print(z + w)z = w4. (2 points) How many different objects will exist in the computer’s memory after the code has been executed?Assume that no optimizations have been performed and that garbage collection has been executed. WHY?

Consider the following code.x = [1]y = xz = x[:]x, y = z, zwhy there are 3 different objects will exist in the computer’s memory after the code has been executed?Assume that no optimizations have been performed and that garbage collection has not been executed.

How does Python manage memory and garbage collection for objects that are no longer referenced in the code?

16. a = [1,2,3]b= aa == ba is bc = list(a)a == ca is cb == cWhich of the following statements is True?a is b: This statement checks if a and b reference the same object in memory, and it returns True.a == c: This statement checks if a and c contain the same elements, and it returns False because they are different objects.b == c: This statement checks if b and c contain the same elements, and it returns False because they reference different objects.a is c: This statement checks if a and c reference the same object in memory, and it returns True because they were created from the same list.

How many objects and reference variables are there for the given Python code?class A:    print("Inside class")A()A()obj=A()

1/1

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.