I2C communication is not working

Question

I have been working on my project but I get an error and I need some advice or correction.

I use I2C method to communicate between master and slave. I've got two questions:

1. I send 7 inputs (from master) to the slave but only receives 6
2. Will my coding make an output like this video

Master Code

// Master Code
#include <Wire.h>
byte nChar =0;
char input[7];
void setup () {
 Serial.begin(9600);
 Wire.begin();
}
void loop()
{
 if (nChar < 7) {if (Serial.available()) { input[nChar++] = Serial.read();}}
 else {
 Wire.beginTransmission(1);
 //a
 Wire.write(input[0]);
 Serial.print("Drawer : ");
 Serial.println(input[0]);
 //x
 Wire.write(input[1]);
 Serial.print("Output : ");
 Serial.println(input[1]);
 //00
 Wire.write(input[2]);
 Wire.write(input[3]);
 Serial.print("Output : ");
 Serial.print(input[2]);
 Serial.println(input[3]);
 //y
 Wire.write(input[4]);
 Serial.print("Output : ");
 Serial.println(input[4]);
 //00
 Wire.write(input[5]);
 Wire.write(input[6]);
 Serial.print("Output : ");
 Serial.print(input[5]);
 Serial.println(input[6]);
 delay(15);
 while (Serial.available())
 {
 //Remove extra
 Serial.read();
 }
 Wire.endTransmission();
 nChar = 0;
 }
 }

Slave Code

#include <Wire.h>
byte ledPins[] = {12, 11, 10, 9, 8, 7, 6};
byte ledPins2[] = {12, 11, 10, 9, 8, 7, 6};
byte count;
byte nChar;
char c;
char input[8];
char data[30] = "";
int n;
#define nBits sizeof(ledPins)/sizeof(ledPins[0])
#define nBits sizeof(ledPins2)/sizeof(ledPins[0]) 
void setup()
{
 Wire.begin(1);
 Serial.begin(9600);
 for (byte i = 0; i < nBits; i++) {
 pinMode(ledPins[i], OUTPUT);
 }
for (byte i = 0; i < nBits; i++) {
 pinMode(ledPins2[i], OUTPUT);
 }
 Serial.begin(9600);
}
void loop()
{
 if (nChar < 8) { //accumulate three characters
 if (Serial.available()) {
 c = Serial.read();
 input[nChar++] = c;
 }
 }
 else {
 input[8] = 1; //atoi() expects string terminator
 n = atoi(input); //convert the input characters to integer
 dispBinary(n);
 delay(100); //wait for any additional characters 
 while (Serial.available()) { //ignore them
 c = Serial.read();
 }
 nChar = 1;
 }
} 
void receiveEvent(int howMany)
{
 while (1 < Wire.available()) {
 char c = Wire.read();
 Serial.print(c);
 }
}
void dispBinary(int n) //Vertical
{
 if (n >= 0 && n <= 5) {
 for (byte i = 0; i < nBits; i++) {
 digitalWrite (ledPins[i], n & 1);
 n /= 2;
 }
 }
}
void dispBinary2(int n) //Horizontal
{
 if (n >= 0 && n <= 5) {
 for (byte i = 0; i < nBits; i++) {
 digitalWrite (ledPins2[i], n & 1);
 n /= 2;
 }
 }
}
void b()
{
 if (data[0] != 'a')
 {
 dispBinary(atoi("00"));
 delay(100);
 dispBinary2(atoi("00"));
 delay(100);
 goto fin;
 }
 if (data[1] = 'x')
 {
 char xdata[2] = {data[2], data[3]};//(data[2], data[3]);
 dispBinary2(atoi(xdata));
 delay(100);
 }
 if (data[4] = 'y')
 {
 char ydata[2] = {data[5], data[6]};//(data[5], data[6]);
 dispBinary(atoi(ydata));
 delay(100);
 }
fin:;
}

• Jot: Where is the Wire.onRequest ?

Answer 1

Most coders would your while (1 < Wire.available()) { write as while ( Wire.available() > 1) { and read it "repeat while there is more the 1 byte available". but wait, why leave that one byte there? you want "repeat while there is no more byte available".

you need while (0 < Wire.available())

• Muhammad Nazirul Na'im Maznan: Which coding sir ?
• Juraj: change while (1 < Wire.available()) to while (0 < Wire.available())
• Muhammad Nazirul Na'im Maznan: Another question why it became on same line. Why not it become one line one output. I cant put image to share with you. How to do ?
• Juraj: replace Serial.print(c); with Serial.println(c); ln is line