Interrupt problem...?

Question

I am working on a code including interrupts. They, however, refuse to work, even in this simple code that could be the most basic example of interrupts.

In this simplified extract I have a momentary push button attached to an Arduino Pro Mini's PIN 2, with the other side grounded. I have this code uploaded:

#include <Encoder.h> // is this library the cause??...
#define PIN_1 4
#define PIN_2 3 //interrupt pin //using these for a rotary encoder, not in this extract
#define PIN_B 2 //interrupt pin // the rotary encoder also works as a push-button
void setup() {
 pinMode(PIN_B, INPUT_PULLUP);
 Serial.begin(9600);
 attachInterrupt(PIN_B, ButtonPressISR, FALLING);
 attachInterrupt(PIN_B, ButtonReleaseISR, RISING);
}
void loop() {
 Serial.println("Don't interrupt me");
 delay(10);
}
void ButtonPressISR() {
 Serial.println("Yay, I'm pressed!");
 delay(1000);
}
void ButtonReleaseISR() {
 Serial.println("Yay, I'm released!");
 delay(1000);
}

One would expect that without action, the Arduino floods the serial console with "Don't interrupt me". And that's correct.

When I press the button, however, nothing on Earth happens, neither of the interrupt service routines starts. No matter how hard I try, "Yay, I'm pressed" and "Yay, I'm released" never appears on the console output, and the 1 s delay never happens.

The circuit, shown below, is tested with a multimeter, and is error-free. Pin 2 does get pulled down, and back, upon button press/release.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

• James Waldby - jwpat7: Generally, Serial.println() does not work correctly from within an ISR, as println() uses interrupts itself
• 001: I believe adding the 2nd interrupt disables the first. Use a single interrupt with CHANGE. Also, digitalPinToInterrupt() might help.
• Neinstein: @JamesWaldby-jwpat7 thanks. But does this explain the lack of the 1 sec delay as well? i.e. does the ISR fails completely as is because of println()?
• Jot: Neinstein, you missed what @JohnnyMopp is saying. Number '2' or '3' are not valid interrupts for the Pro Mini. The function digitalPinToInterrupt is preferred. arduino.cc/en/Reference/AttachInterrupt (read the whole page). No one sets two interrupts for a single pin, it does not work, a CHANGE does work. Set a volatile byte in the ISR and use that byte in the loop. If you have an encoder, why not use the Encoder library that you have already downloaded ? pjrc.com/teensy/td_libs_Encoder.html
• Neinstein: @jot the Encoder libary does not support buttons AFAIK.

Answer 1

Neither delay or print should be used in an interrupt routine as they use interrupts themselves, instead set a flag (usually a volatile variable) and test for that in the loop. Also be aware that due to contact bounce the interrupt will trigger multiple times for each button press.

Answer 2

They, however, refuse to work,

more accurately, you are unable to make them work.

1. don't delay large amount of time;
2. be careful with calling routines that are also called outside of the isr;
3. one isr for one pin.