Arduino send bad signal to interrupt pin - arduino

Question

I have connected coin hopper and coin acceptor to one arduino uno, coin acceptor connected to pin 2, coin hopper to pin 3 - sensor and pin 7 - relay. When coin hopper switch relay, it is executing coininterrupt
for coin hopper I am using this script link
coin acceptor script: link
I need this 2 scripts working on 1 arduino
my code:
#define SENSOR 3
#define RELAY 7
#define ACCEPTOR 2
volatile boolean insert = false;
int pulse=0,count;
char sen;
int temp=0;
unsigned long int timer;
void setup()
{
Serial.begin(9600);
pinMode(SENSOR,INPUT_PULLUP);
pinMode(RELAY,OUTPUT);
sen=digitalRead(SENSOR);
digitalWrite(RELAY, HIGH);
attachInterrupt(digitalPinToInterrupt(ACCEPTOR), coinInterrupt, RISING);
}
void loop()
{
if (insert) {
insert = false;
Serial.println("coin");
delay(1000);
}
if(Serial.available())
{
timer=millis();
// temp is amount to dispense send to arduino
temp=Serial.parseInt();
if(temp>0){
digitalWrite(RELAY,LOW);}
}
sen=(sen<<1)|digitalRead(SENSOR);
// if hopper sensor read drop coin
if(sen==1)
{
timer=millis();
pulse++;
sen&=0x03;
Serial.println("out 1");
//if dispensed coins equal with coins to dispense stop engine
if(pulse==temp)
{
digitalWrite(RELAY,HIGH);
pulse=0;
temp=0;
}
}
// if amount dispensed is not equal with amount to dispense and engine running, stop
if((digitalRead(RELAY)==LOW)&(millis()-timer>2000))
{
digitalWrite(RELAY,HIGH);
pulse=0;
temp=0;
}
}
void coinInterrupt() {
insert = true;
}
I was trying to change pins (arduino uno support interrupts on pin 2 and 3 only) but problem still appears so I guess there is issue in the code

Answer 1

your sketch does not run in this state :
first fix errors :
declare insert as volatile
remove cpulse (not used anywhere)
change 'if()' to (I suppose) 'if (insert) ....'
remove stuff with 'sen' var : simply use if(digitalRead(SENSOR)) or if(!digitalRead(SENSOR))
except if you need to store relay state.
use logical operators like || or && unless you really need bitwise operations
example of result sketch :
#define SENSOR 3
#define RELAY 7
volatile boolean insert = false;
byte amountToDispense = 0;
int pulse = 0;
int temp = 0;
unsigned long int timer;
void setup()
{
Serial.begin(9600);
pinMode(SENSOR, INPUT_PULLUP);
pinMode(RELAY, OUTPUT);
digitalWrite(RELAY, HIGH);
attachInterrupt(digitalPinToInterrupt(2), coinInterrupt, RISING);
}
void loop()
{
if (insert ) {
insert = false;
Serial.println("coin");
delay(1000);
}
if (Serial.available())
{
timer = millis();
temp = Serial.parseInt();
if (temp > 0) {
//amountToDispense = Serial.read() - 48;
digitalWrite(RELAY, LOW);
}
}
if (digitalRead(SENSOR))
{
timer = millis();
pulse++;
Serial.println("out 1");
if (pulse >= temp)
{
digitalWrite(RELAY, HIGH);
pulse = 0;
temp = 0;
}
}
if (!digitalRead(RELAY) && (millis() - timer > 2000))
{
digitalWrite(RELAY, HIGH);
pulse = 0;
temp = 0;
}
}
void coinInterrupt() {
insert = true;
}

Answer 2

What is this supposed to do?
sen=(sen<<1)|digitalRead(SENSOR);
You init sen with digitalRead(SENSOR);
Assuming that pin is LOW when you start the sketch and turns HIGH, sen will become 1.
Next you do sen &= 0x03 so sen is still 1.
Again sen=(sen<<1)|digitalRead(SENSOR); , sen will either be 2 or 3.
Next loop run sen=(sen<<1)|digitalRead(SENSOR); sen is now 4 or 6. and so on...
I don't have time to think about what you want to achieve but this is definitely a problem as you'll only enter if (sen == 1) once and never again.
If this is not sufficient you should probably improve your post as it is unclear what arduino sends bad signal to interrup pin is supposed to mean. That doesn't make sense. Explain the expected behaviour of your program and how it behaves instead. add more comments so it becomes clear what you intend to do with each block of code so we don't have to interpret