Arduino Error 1 expected unqualified-id before 'else'

Question

Arduino Error: Where have i gone wrong in this simple and extremely small amount of code to receive the above error message? I cannot figure it out.

int ledPin = A0;
int bumpPin = A1;
void setup() {
pinMode(ledPin, OUTPUT);
pinMode(bumpPin, INPUT);
}

void loop() {
digitalRead(bumpPin);
if (bumpPin == HIGH);
digitalWrite(ledPin,HIGH);
}else{
digitalWrite(ledPin,LOW);
}

I'd say you need a [good book](https://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list). (Also, why did you tag this as Java?) — Rakete1111, Oct 01 '17 at 16:09

Steampunkery · Accepted Answer · 2017-10-01T17:13:57.820

1

Definitely read a tutorial on C++. What you have here is a basic syntax error. An if/else statements uses the following syntax:

if (condition) {
    // Do stuff here
} else {
    // Do other stuff here
}

You have a semi colon after your condition in your if statement. Change that to a curly brace and you’re good! So this:

if (bumpPin == HIGH);

Should be this:

if (bumpPin == HIGH) {

Here is an online C++ tutorial.

Here is that tutorial’s section on if statements.

edited Oct 01 '17 at 17:13

answered Oct 01 '17 at 16:14

Steampunkery

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Tried that with the error: 'else' without a previous 'if' – Steve Oct 01 '17 at 16:18
No, don't read a tutorial on _C/C++_ because that language does not exist. Read a [good book](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) or adequate tutorial on C++. – Captain Obvlious Oct 01 '17 at 16:37
Yes it did end up working missing a brace at the end of the code thanks – Steve Oct 02 '17 at 17:00
Glad to be of help! – Steampunkery Oct 02 '17 at 17:19
`bumpPin == HIGH` will always be `false`. – gre_gor Oct 04 '17 at 18:48

Arduino Error 1 expected unqualified-id before 'else'

1 Answers1