OAU CSC 201 Practice Quiz

In the code shown above, we have used a decorator in order to avoid the Zero Division Error. Hence the output of this code is:

    hello
    NO

In Python, to define a block of code we use indentation. Indentation refers to whitespaces at the beginning of the line.

The code shown above returns a new list containing only those elements of the list l which do not amount to zero. Hence the output is: [1, 2, ‘hello’].

9 isn’t allowed in an octal number.

In order to add ‘n’ blank spaces after a given string ‘S’, we use the formatting option:(“%-ns”%S).

The binary form of 1 is 0001. The expression x<<2 implies we are performing bitwise left shift on x. This shift yields the value: 0100, which is the binary form of the number 4.

The expression shown above does not result in an error. The output of this expression is -%5d0 989. Hence this statement is incorrect.

String x is not being altered and i is in x[:-1].

The code shown above is used to swap the contents of two memory locations using bitwise X0R operator. Hence the output of the code shown above is: 20 10.

0O11 is an octal number.

^ is the XOR operator. The binary form of 4 is 0100 and that of 12 is 1100. Therefore, 0100^1100 is 1000, which is equal to 8.

Python has a nifty feature called documentation strings, usually referred to by its shorter name docstrings. DocStrings are an important tool that you should make use of since it helps to document the program better and makes it easier to understand.

[::-1] reverses the list.

The function not returns true if the argument amounts to false, and false if the argument amounts to true. Hence the first function returns false, and the second function returns false.

The code shown above prints 1234 for the format specified %d, ‘1234  ’ for the format specifier %-6d (minus ‘-‘ sign signifies left justification), and 001234 for the format specifier %06d. Hence the output of this code is: ‘integers:…1234…1234  …001234’

“i” is not in “abcdef”.

The symbol ‘&’ represents bitwise AND. This gives 1 if both the bits are equal to 1, else it gives 0. The binary form of 15 is 1111 and that of 12 is 1100. Hence on performing the bitwise AND operation, we get 1100, which is equal to 12.

The formatting method {1:<10} represents the second positional argument, left justified in a 10 character wide field.

The code shown above rounds the floating point value to two decimal places. In this code, a variety of addition formatting features such as zero padding, total field width etc. Hence the output of this code is: ‘3.35 | 03.35 | +003.3’.

^ is the Binary XOR operator.

Functions are reusable pieces of programs. They allow you to give a name to a block of statements, allowing you to run that block using the specified name anywhere in your program and any number of times.

The one’s complement of a value is obtained by simply changing all the 1’s to 0’s and all the 0’s to 1’s. Hence the one’s complement of 110010101 is 001101010.

The code shown first adds the element ‘san’ to the set z. The set z is then updated and two more elements, namely, ‘p’ and ‘q’ are added to it. Hence the output is: {‘a’, ‘b’, ‘c’, ‘p’, ‘q’, ‘san’}

Let’s break it down step by step:

Code:

z = set(‘abc’)                        # Create a set with elements from the string ‘abc’

z.add(‘san’)                          # Add the string ‘san’ to the set

z.update(set([‘p’, ‘q’]))             # Add elements ‘p’ and ‘q’ from another set to `z`

z                                     # Display the final set

Step-by-Step Explanation:

Step 1: Create a set

z = set(‘abc’)

• The string ‘abc’ is turned into a set of individual characters:

z = {‘a’, ‘b’, ‘c’}

• Note: Sets do not allow duplicate elements, so each character in ‘abc’ is added only once.

Step 2: Add the string ‘san’

z.add(‘san’)

• The .add() method adds a single element to the set.

• ‘san’ is treated as a single item, not split into characters.

• The set now looks like this:

z = {‘a’, ‘b’, ‘c’, ‘san’}

Step 3: Update the set with set([‘p’, ‘q’])

z.update(set([‘p’, ‘q’]))

• The .update() method adds multiple elements from another set or iterable to the set.

• The elements ‘p’ and ‘q’ are added individually to z.

• The set now looks like this:

z = {‘a’, ‘b’, ‘c’, ‘san’, ‘p’, ‘q’}

Step 4: Display the set

z

• The set contains all unique elements added in previous steps:

{‘a’, ‘b’, ‘c’, ‘san’, ‘p’, ‘q’}

Important Notes:

1. Sets do not allow duplicate values: If an element already exists, adding it again has no effect.

2. Order of elements in a set is unpredictable: Sets are unordered collections, so the display order might differ.

Correct Answer:

c) {‘a’, ‘b’, ‘c’, ‘p’, ‘q’, ‘san’}

Case is always significant while dealing with identifiers in python.

The binary form of 1 is 0001. The expression x<<2 implies we are performing bitwise left shift on x. This shift yields the value: 0100, which is the binary form of the number 4.

The code shown above demonstrates rebinding using a static method. This can be done with or without a decorator. The output of this code will be 100.

The line of code can be translated to state that ‘f’ is printed if the argument passed to the Boolean function amount to zero. Else ‘t’ is printed. The argument given to the Boolean function in the above case is ‘spam’, which does not amount to zero. Hence the output is t.

In most cases the value of two’s complement is different from the original value. However, there are cases in which the two’s complement value may be equal to the original value. For example, the two’s complement of 10000000 is also equal to 10000000. Hence the statement is false.

The code shown above shows a general decorator which can work with any number of arguments.

range(int(2.0)) is the same as range(2).

Variable names can be of any length.

The string x is being shortened by one character in each iteration.

Any odd number on being AND-ed with 1 always gives 1. Any even number on being AND-ed with this value always gives 0.

Since we have specified that the order of the output be: {a}, {0}, {abc}, hence the value of associated with {a} is printed first followed by that of {0} and {abc}. Hence the output of the code shown above is: ‘2.5, 10, [1, 2]’.

SyntaxError, True is a keyword and it’s value cannot be changed.

The function max() is being used to find the maximum value from among -3, -4 and false. Since false amounts to the value zero, hence we are left with min(0, 2, 7) Hence the output is 0 (false).

Suppose we have an expression ~A. This is evaluated as: -A – 1. Therefore, the expression ~100 is evaluated as -100 – 1, which is equal to -101.

Addition and Subtraction” are at the same precedence level. Similarly, “Multiplication and Division” are at the same precedence level. However, Multiplication and Division operators are at a higher precedence level than Addition and Subtraction operators.

Execute in the shell.

Variable x is incremented and printed twice.

The given number is -122. Here the total number of digits (including the negative sign) should be 6 according to the expression. In addition to this, there is a negative sign in the given print expression. Hence the output will be −−00122.