Question

Assignment (C language)

We will simulate the status of 8 LEDs that are connected to a microcontroller. Assume that the state of each LED (ON or OFF) is determined by each of the bits (1 or 0) in an 8-bit register (high-speed memory).

1. Declare a char variable called led_reg and initialize it to 0.
   Assume that the least-significant bit (lsb) controls LED#0 and the most-significant bit (msb) controls LED#7.
   
2. In the main function, build and present a menu to the user and enclose it in a loop that ends when the Quit option is chosen. Use if-else or switch-case statements to implement the operations described:
   
   Menu:
   0. Turn ON LED#4
   1. Turn OFF LED#4
   2. Add 1 to the LED Register
   3. Turn ON all the even numbered LEDs
   4. Turn OFF all the even numbered LEDs
   5. Toggle the state of LED#3.
   6. Shift the values of the LED Register right by 1
   7. Quit
   
3. Create a function called: printBinary() with the following properties:
   
   • one input argument of type char
   • returns nothing
   • prints the binary representation (in 8-bits) of the char input value.
   
4. Use the printBinary() function to display the value of the LED register before the menu.

HINTS:

• Each operation can be performed with a single expression (line of code).
• Append 0b or 0x to numerical constants to denote binary or hexadecimal values in source code, respectively.
  10101010₂ = 0b10101010
  FF₁₆ = 0xFF

Answer 1

Here is the code:

#include <bits/stdc++.h>

using namespace std;
void printBinary(char);

int main()
{
   char led_reg = '0';
   while(1){
       printBinary(led_reg);
       cout<<"\n0. Turn ON LED #4 .\n";
       cout<<"1. Turn OFF LED #4 .\n";
       cout<<"2. Add 1 to the LED Register .\n";
       cout<<"3. Turn ON all the even numbered LEDs .\n";
       cout<<"4. Turn OFF all the even numbered LEDs .\n";
       cout<<"5. Toggle the state of LED #3 .\n";
       cout<<"6. Shifte the values of the LED Register right by 1 .\n";
       cout<<"7. Quit.\n";
       int choice;
       cin>>choice;
       switch(choice)
       {
           case 0:
               led_reg = (led_reg | (1 << 2));            // All the bits that we count are from left side to right
               break;                                               //but the operations are performed in opposite dxn
           case 1:                                                   // so we do 7-i-1 for the ith bit from left
               led_reg = (led_reg & ~(1 << 2));        //hence we use 2 for 4 as 7-4-1=2
               break;
           case 2:
               led_reg++;
               break;
           case 3:
               for(int i = 0;i<8; i++){
                   if(i%2==0){
                       led_reg = (led_reg | (1 << i));
                       }
                   }
               break;
           case 4:
               for(int j = 0;j<8; j++){
                   if(j%2==0){
                       led_reg = (led_reg & ~(1 << j));
                       }
                   }
               break;
           case 5:
               led_reg = (led_reg ^ (1 << (5)));
               break;
           case 6:
               led_reg = led_reg>>1;
               break;
           case 7:
               exit(0);
               break;

           default:
               cout<<"Invalid Choice!!\n Try agin\n";
       }
   }
    return 0;
}

void printBinary(char c){
   /* step 1 */
    if (c > 1)
        printBinary(c/2);

    /* step 2 */
    cout << c % 2;
   }