Only return button state once in X amount of time?

Question

I want to detect if a button has been pressed, but what I don't want is for it to keep returning the "pressed" state more than once within a second or two.

For example, with this code:

void loop() {
 passButtonState = digitalRead(passButton);
 if (passButtonState == LOW) {
 Serial.println("PASSED!");
 }
}

Would spit out something like this with a single quick press of the button:

PASSED!
PASSED!
PASSED!
PASSED!
PASSED!
PASSED!
PASSED!
PASSED!
PASSED!

But what I want is for a single button press to just return PASSED! once for that single button press.

Answer 1

To avoid this problem you can use a boolean variable as a flag that permit to enter inside the if only once.

bool flag = true;
void loop() {
 passButtonState = digitalRead(passButton);
 if (passButtonState == LOW && flag) {
 Serial.println("PASSED!");
 flag = false;
 }
}

• Shpigford: But this permanently disables that button from then on. I just want it to only return once for each press as opposed to a dozen.
• rebatoma: @Shpigford you can reset the flag = true after a period of time.

Answer 2

If I understand the problem correctly, I think the answer might be a pretty common one. Your circuit may be suffering from "bounce" and the solution is called "debounce."

Check out this link for a possible solution: Software Debounce

Answer 3

void loop() {
 passButtonState = digitalRead(passButton);
 if (passButtonState == LOW) {
 Serial.println("PASSED!");
 while (passButtonState == LOW) delay(1);
 }
}

Answer 4

You can add a delay.

void loop() {
 passButtonState = digitalRead(passButton);
 if (passButtonState == LOW) {
 Serial.println("PASSED!");
 delay(1000);
 }
}

• aaa: Could you explain why this would be a good solution? While it may be trivial to you and me, it certainly won't be to everyone. Also, you should state that adding a delay will "slow" the whole program. You might even give alternative solutions.