Answer:
All of these statements are true.
Explanation:
Since the while loop is reversing the integer number and leaving the highest order digit in the num and stores the reversed number in the newNum variable.
It skips one digit so if the num is in the range of [100,1000] it will result in a number between 10 and 100.
This loop can never go in infinite loop for any initial value of num because the loop will run as many times as the number of digits.
and if the value of the num is <=10 the while loop will never run and the value of newNum will be 0.
Thus is true
Encryption helps to keep data virtually safe
Answer:
The solution code is written in Python 3.
- def processCSV(CSV_string, index):
- data = CSV_string.split(",")
- result = data[index]
- return int(result)
-
- processCSV('111,22,3333,4', 2)
Explanation:
Firstly, create function and name it as <em>processCSV()</em> with two parameters, <em>CSV_string</em> and <em>index</em> (Line 1).
Since the CSV_string is a collection of integers separated by comma, we can make use of Python String built-in method <em>split() </em>to convert the <em>CSV_string </em>into a list of numbers by using "," as separator (Line 2).
Next, we can use the <em>index</em> to extract the target value from the list (Line 3).
Since the value in the list is still a string and therefore we shall use Python built-in method int() to convert the target value from string type to integer and return it as output of the function (Line 4)
We can test our function by using the sample test case (Line 6) and we shall see the expected output: 3333.
Flip bit position 5 to accomplish this. This maps to hex value 0x20, where the least significant bit is assumed to be at position 0.
Example: ascii "A" = 0x41, "a" = 0x61. 0x41 xor 0x61 = 0x20.
You would implement a flip function by XOR'ing the character value with 0x20.
Answer:
Check the explanation
Explanation:
223.1.17/24 indicates that out of 32-bits of IP address 24 bits have been assigned as subnet part and 8 bits for host id.
The binary representation of 223.1.17 is 11011111 00000001 00010001 00000000
Given that, subnet 1 has 63 interfaces. To represent 63 interfaces, we need 6 bits (64 = 26)
So its addresses can be from 223.1.17.0/26 to 223.1.17.62/26
Subnet 2 has 95 interfaces. 95 interfaces can be accommodated using 7 bits up to 127 host addresses can represented using 7 bits (127 = 27)
and hence, the addresses may be from 223.1.17.63/25 to 223.1.17.157/25
Subnet 3 has 16 interfaces. 4 bits are needed for 16 interfaces (16 = 24)
So the network addresses may range from 223.1.17.158/28 to 223.1.17.173/28
