Exam Code : PCEP-30-02
Exam Name : PCEP - Certified Entry-Level Python Programmer
Vendor Name :
"Python"
PCEP-30-02 Dumps
PCEP-30-02 Braindumps PCEP-30-02 Real Questions PCEP-30-02 Practice Test PCEP-30-02 Actual Questions
PCEP - Certified Entry-Level Python Programmer
https://killexams.com/pass4sure/exam-detail/PCEP-30-02
Consider the following code snippet: w = bool(23)
x = bool('')
y = bool(' ')
z = bool([False])
Which of the variables will contain False?
z
x
y
w
Explanation:
Topic: type casting with bool() Try it yourself:
print(bool(23)) # True print(bool('')) # False print(bool(' ')) # True print(bool([False])) # True
The list with the value False is not empty and therefore it becomes True The string with the space also contain one character
and therefore it also becomes True
The values that become False in Python are the following: print(bool('')) # False
print(bool(0)) # False print(bool(0.0)) # False print(bool(0j)) # False print(bool(None)) # False print(bool([])) # False print(bool(())) # False print(bool({})) # False print(bool(set())) # False print(bool(range(0))) # False
What is the expected output of the following code? def func(num):
res = '*'
for _ in range(num): res += res
return res
for x in func(2): print(x, end='')
**
The code is erroneous.
*
****
Explanation:
Topics: def return for Try it yourself:
def func(num): res = '*'
for _ in range(num): res += res
return res
for x in func(2): print(x, end='') # ****
# print(x, end='-') # *-*-*-*- print()
print(func(2)) # ****
The for loop inside of the function will iterate twice. Before the loop res has one star.
In the first iteration a second star is added. res then has two stars.
In the second iteration two more stars are added to those two star and res will end up with four stars. The for loop outside of the function will just iterate through the string and print every single star. You could get that easier by just printing the whole return value.
What is the expected output of the following code? num = 1
def func():
num = num + 3 print(num) func()
print(num)
4 1
4 4
The code is erroneous.
1 4
1 1
Explanation:
Topics: def shadowing Try it yourself:
num = 1 def func():
# num = num + 3 # UnboundLocalError: ... print(num)
func() print(num) print('----------')
num2 = 1 def func2():
x = num2 + 3 print(x) # 4 func2()
print('----------')
num3 = 1 def func3():
num3 = 3 # Shadows num3 from outer scope print(num3) # 3
func3()
A variable name shadows into a function. You can use it in an expression like in func2()
or you can assign a new value to it like in func3()
BUT you can not do both at the same time like in func()
There is going to be the new variable num
and you can not use it in an expression before its first assignment.
The result of the following addition: 123 + 0.0
cannot be evaluated
is equal to 123.0
is equal to 123
Explanation:
Topics: addition operator integer float Try it yourself:
print(123 + 0.0) # 123.0
If you have an arithmetic operation with a float, the result will also be a float.
What is the expected output of the following code? print(list('hello'))
None of the above.
hello
[h, e, l, l, o]
D. ['h', 'e', 'l', 'l', 'o']
E. ['h' 'e' 'l' 'l' 'o']
Try it yourself:
print(list('hello')) # ['h', 'e', 'l', 'l', 'o']
A string is a sequence of characters
and works very fine with the list() function.
The result is a list of strings, in which every character is a string of its own.
What is the default return value for a function that does not explicitly return any value?
int
void
None
Null
public
print(func1()) # None def func2():
return
print(func2()) # None
If a function does not have the keyword return the function will return the value None The same happens if there is no value after the keyword return
Which of the following lines correctly invoke the function defined below: def fun(a, b, c=0):
# Body of the function. (Select two answers)
fun(0, 1, 2)
fun(b=0, a=0)
fun(b=1)
fun()
Explanation:
Topics: functions positional parameters keyword parameters Try it yourself:
def fun(a, b, c=0):
# Body of the function. pass
fun(b=0, a=0) fun(0, 1, 2)
# fun() # TypeError: fun() missing 2 required
# positional arguments: 'a' and 'b'
# fun(b=1) # TypeError: fun() missing 1 required
# positional argument: 'a'
Only the parameter c has a default value. Therefore you need at least two arguments.
What is the expected output of the following code? x = '''
print(len(x))
1
2
The code is erroneous.
0
Explanation:
Topics: len() escaping Try it yourself: print(len(''')) # 1
The backslash is the character to escape another character.
Here the backslash escapes the following single quote character. Together they are one character.
Which of the following statements are true? (Select two answers)
The ** operator uses right-sided binding.
The result of the / operator is always an integer value.
The right argument of the % operator cannot be zero.
Addition precedes multiplication.
Explanation:
Topics: exponentiation/power operator right-sided binding Try it yourself:
print(4 ** 3 ** 2) # 262144
print(4 ** (3 ** 2)) # 262144
print(4 ** 9) # 262144
print(262144) # 262144
# print(7 % 0) # ZeroDivisionError
If you have more than one power operators
next to each other, the right one will be executed first. https://docs.python.org/3/reference/expressions.html#the-power- operator To check the rest of a modulo operation,
Python needs to divide the first operand by the second operand. And like in a normal division, the second operand cannot be zero.
What do you call a tool that lets you lanch your code step-by-step and inspect it at each moment of execution?
A debugger
An editor
A console
Explanation:
Topic: debugger https://en.wikipedia.org/wiki/Debugger
What is the expected output of the following code? list1 = [1, 3]
list2 = list1 list1[0] = 4 print(list2)
A. [1, 4]
B. [4, 3]
C. [1, 3, 4]
D. [1, 3]
Explanation:
Topics: list reference of a mutable data type Try it yourself:
list1 = [1, 3] list2 = list1 list1[0] = 4
print(list2) # [4, 3]
print(id(list1)) # e.g. 140539383947452
print(id(list2)) # e.g. 140539383947452 (the same number) A list is mutable.
When you assign it to a different variable, you create a reference of the same object. If afterwards you change one of them, the other one is changed too.
How many stars will the following code print to the monitor? i = 0
while i <= 3: i += 2
print('*')
one
zero
two
three
while i <= 3: # i=0, i=2 i += 2
print('*')
"""
*
* """
In the first iteration of the while loop i is 0 i becomes 2 and the first star is printed.
In the second iteration of the while loop i is 2 i becomes 4 and the second star is printed.
i is 4 and therefore 4 <= 3 is False what ends the while loop.
What is the expected output of the following code if the user enters 2 and 4? x = input()
y = input() print(x + y)
4
6
24
2
Explanation:
Topics: input() plus operator string concatenation Try it yourself:
# x = input()
# y = input()
x, y = '2', '4' # Just for convenience print(x + y) # 24
As always the input() function return a string.
Therefore string concatenation takes place and the result is the string 24