Author: Christian Lerche
Date: 05-12-2009
MCU: LPC1768
Notes: Simple graycode upcounter, using onboard LEDs.

*/

#include "mbed.h"

BusOut LEDs(LED4, LED3, LED2, LED1);        // Make 4 bit bus with LEDs
unsigned char i;                            // Char to read for LEDs

int main() {                                // Main structure
    i=0;                                    // initialize i to 0
    while(1) {                              // While-loop (Do forever)
        LEDs=i^(i>>1);                      // This converts the binary to graycode
        wait(2);                            // Wait for .5 seconds
        i=i+1;                              // Every loop, add one to i
        if(i==16) {                         // If i is 16, make i 0
            i=0;                            // This makes it 0
        }                                   // End of if
    }                                       // End of while-loop (Do not anymore)
}                                           // End of main structure

LEDs=i^(i>>1);

This is a simple 4 bit up counter. <<code>> Author: Christian Lerche Date: 05-12-2009 MCU: LPC1768 Notes: Simple graycode upcounter, using onboard LEDs. */ #include "mbed.h" BusOut LEDs(LED4, LED3, LED2, LED1); // Make 4 bit bus with LEDs unsigned char i; // Char to read for LEDs int main() { // Main structure i=0; // initialize i to 0 while(1) { // While-loop (Do forever) LEDs=i^(i>>1); // This converts the binary to graycode wait(2); // Wait for .5 seconds i=i+1; // Every loop, add one to i if(i==16) { // If i is 16, make i 0 i=0; // This makes it 0 } // End of if } // End of while-loop (Do not anymore) } // End of main structure <</code>> Please can someone explain what the following statements actually do: <<code>> LEDs=i^(i>>1); <</code>> What does the '^' hat mean? As I understand it 'i>>1' converts the binary to graycode But I have also seen 'i>>2' and 'i>>3' Does this also convert it to graycode? Cheers David

The '^' hat means XOR (exclusive OR). This operation results in inverting the source bit for every digit position that has a '1' in the maskpattern. XOR truth table: 0 ^ 0 = 0 1 ^ 0 = 1 0 ^ 1 = 1 1 ^ 1 = 0 The 'i>>1' is a 1 position bitshift to the right. This results in a value that is divided by 2. The resulting sequence for i=0 to i=15 is: (0000) ^ (0000) = 0000; (0001) ^ (0000) = 0001; (0010) ^ (0001) = 0011; (0011) ^ (0001) = 0010; ... (1110) ^ (0111) = 1001; (1111) ^ (0111) = 1000; That results in a Graycode. Graycodes change only one bit between sequential values.