Write a single C++ statements to accomplish each of the following tasks:

1. In one statement, assign the sum of the current value of x and y to z and postincrement the value of x.
   1. Determine whether the value of the variable count is greater than 10. If it is, print

“Count is greater than 10.”.

3. Pre-decrement the variable of x by 1, then subtract it from the variable total.
4. Multiply variable power by x and assign the result to power.  

(b)  

1. Given a=3, b=2, and c=5, evaluate the final value of R for the following statements:
2. 1. R = ++a * c++ % b + (a * b) > 20
   2. R = b * c / a++ + ++a + ++b
   3. R = (a * b * (c + (9 * 3 / 3)))
   4. R = b % b + b * b – b / b
   5. R = c / b * 10 % 3 < 5

SHOW THE STEPS.

(c) Write a program that prompt the user to enter a string. Encrypts a message where each of the character in the string will be replaced by the next 3 letters of the current letter. For example: if the user input a message “Hello”, then the encrypted message is “Khoor”.

Sample Output Run

Enter a String : My Best Friend

Encrypted message is : P|#Ehw#Iuhqg

Answer:

a. In one statement, assign the sum of the current value of x and y to z and post-increment the value of x:

int z = (x++) + y;

i. Determine whether the value of the variable count is greater than 10. If it is, print “Count is greater than 10.”:

int count = 20;

cout << ((count > 10) ? "Count greater than 10" : "Count less than 10");

iii. Pre-decrement the variable of x by 1, then subtract it from the variable total:

(total - --x);

(b)

Given a=3, b=2, and c=5

Pre-increment:

a = a + 1;

(++a) means, cout << a; will give "4".

Post-increment:

a=a;

a = a + 1;

(a++) means, cout << a; will give "3". If again, cout << a; - it will print 4.

i. R = ++a * c++ % b + (a * b) > 20:

= 4 * 5 % 2 + (4 * 2) > 20

= 4 * 5 % 2 + 8 > 20 [Paranthesis comes 1st]

= 20 % 2 + 8 > 20 [Multiplication, Division, Modulus are all of the same priority. So, we will traverse from left to right]

= 0 + 8 > 20

= 8 > 20 = 0 [Yes = 1 and No = 0]

Answer: 0

[After above operation, a = 4, b = 2, c = 6]

ii. R = b * c / a++ + ++a + ++b:

= 2 * 6 / 4 + 6 + 3 ["a" was 4, a++ gave 4 at that instant and changing "a" to 5, ++a gave 6 at that instant]

= 12 / 4 + 6 + 3

= 3 + 6 + 3 = 12

Answer: 12

[After above operation, a = 6, b = 3, c = 6]

iii. R = (a * b * (c + (9 * 3 / 3)))

= (6 * 3 * (6 + (9 * 3 / 3)))

= (6 * 3 * (6 + 9)) [9 * 3 / 3 = 27 / 3 = 9]

= (6 * 3 * 15) = 270

Answer: 270

[After above operation, a = 6, b = 3, c = 6]

iv. R = b % b + b * b – b / b

= 3 % 3 + 3 * 3 - 3 / 3

= 0 + 9 - 1 = 8

Answer: 8

[After above operation, a = 6, b = 3, c = 6]

v. R = c / b * 10 % 3 < 5

= 6 / 3 * 10 % 3 < 5

= 2 * 10 % 3 < 5

= 20 % 3 < 5

= 2 < 5, gives 1 [Yes = 1 and No = 0]

Answer: 1

c.

#include <iostream>
#include <string>
#include <cctype>
#include <stdlib.h>

using namespace std;

int main()
{
string str, resultString = "";
cout << "Enter a string: ";
getline(cin, str);
  
for(int i = 0; i < str.length(); i++)
{
if(str[i] == ' ')
str[i] = '#';
else
str[i] += 3;
}
 
cout << "\nEncrypted message is: " << str << endl;
 return 0;
}