having some trouble with an if then statement triggered by a High reading on a digital pin/switch

Posted by: @fastrunner08

My program is stuck in live mode even when the pin is reading a low. 

 here is the portion of code I have.  I'm using a TFT screen to display the output.  the code below is in the void loop.  this is the only code in the program that affects these variables so i'm not going to post the rest of the program, since this is already a super long post. 

speed mode is an int, speedmode1 is a string so i can print it to the tft. 

if (speedmode = 0) {

dummymode ();
myGLCD.print(("DummyMode"),10, 200);
}

if (speedmode = 1){                    --- this is getting triggered even if the switch is off. 
livemode();
myGLCD.print(("livemode"),10,200);
}

You are using an arithmetic operator (equals aka "=") where a logical operator (equals aka "==") is required.

So where you say "speed mode = 0)" it actually assigns the value 0 to speed mode, similarly it sets the value to 1 in the following test.

change the lines to 

if (speedmode == 0) {

and 

if (speedmode == 1) {

And try it again.

@will Thank you so much!!!!!  What a simple and stupid mistake.  My programing is rusty, and apparently my frustration wasn't helping me think clearly. 

I really appreciate it.   

@fastrunner08 

It's not stupid, it's extremely common. That's why it's easy to spot 🙂

I'm glad you're back on track.

Posted by: @fastrunner08

@will Thank you so much!!!!!  What a simple and stupid mistake.  My programing is rusty, and apparently my frustration wasn't helping me think clearly. 

I really appreciate it.   

Been there, done that, bought the t-shirt many, many times! You are not alone! 😀 

Honestly, I was zero'ed in on the harder part of the build to the point I overlooked the simple stuff.  I probably will make a post show casing this lovely gadget when i'm done.  it goes  hand in hand with an old john deere computrak 200 planter monitor.  To get seed population they used a radar sensor that was mounted to the planter, and my neighbor bought the planter that was missing it's radar sensor.  Deere wanted something like $2900 for it, and I told him i could find a better/cheaper solution, so i have been working on this.  The hard part i was focusing on was getting the pwm from Arduino just right going into the signal pin of the deere monitor so that the speed it read from the PWM was accurate.  

@fastrunner08

Posted by: @fastrunner08

@will Thank you so much!!!!!  What a simple and stupid mistake.  My programing is rusty, and apparently my frustration wasn't helping me think clearly. 

I really appreciate it.   

We had a discussion on this silent semantic error not too long ago.

When programming, if you can get into the habit of putting the literal to the left hand side of the operator, the compiler won't let you get away with it, because you can't assign to a literal - It will stop compiling and therefore alert you to the error immediately:

  // Will not compile...
  if (1 = speedmode) { 

   }

Cheers

@frogandtoad I don't recall ever seeing that, it's bloody brilliant! Our schools and textbooks need to change how they teach.

@zander

Posted by: @zander

@frogandtoad I don't recall ever seeing that, it's bloody brilliant! Our schools and textbooks need to change how they teach.

It's a tip I picked up over the years - Simple yet effective.

Cheers

@frogandtoad So if i'm understanding you correctly.  you are saying that in a comparison statement like this, you can put the literal to the right or left?  ie. 1 == speedmode or speedmode == 1 would work the same way, but by doing 1 == speedmode the compiler will catch the error if one of the = is missing?

Posted by: @fastrunner08

@frogandtoad So if i'm understanding you correctly.  you are saying that in a comparison statement like this, you can put the literal to the right or left?  ie. 1 == speedmode or speedmode == 1 would work the same way, but by doing 1 == speedmode the compiler will catch the error if one of the = is missing?

 

That's exactly what he is saying. 👍 

@fastrunner08 

Since "a equals b" implies "b equals a", the order of the values tested doesn't matter. So "b==1" and "1==b" are both meaningful and will always return the same result of the test

However, you can only assign a value to a variable, not to a literal value, so

b = 1; makes sense because we're filling the hole known as b with the value 1 but

1 = b is nonsense.

You can't assign a new value to a literal because "1" already has a distinct literal value that can't be modified. 

Note that the compiler should also prevent you from assigning to a const value so:

const  int  myConst = 5;

if (const=6)

should also result in a nastygram from the compiler.

@will Good 2nd point Will. This is why it is important to call a variable by it's full nbame so to speak when declaring them. If it is in fact a constant then declaring it 'const' will stop you from making a mistake.

@fastrunner08 A good way to code it is simply:

if ( speedMode ) {

true condition here

}

else {

false here

}

Using:

if ( ! speedMode ) returns true when the condition if false

In the c language, any non-zero value is "true" and integer zero is false. Doesn't matter the data type of what you're testing. If you have a character value and test

if (myCharHere)

You will get true unless the value has been assigned to zero. Not the character representing zero, the number zero, like:

myCharhere = 0;

Don't use the == operator unless you need to check for a specific value, other than true or false. For example you might use if (myChar == 'j')  This tests for the specific character 'j'

If you need to check a specific integer you can use if ( i == 99 ) Now the test is false for all values of i except 99

When the value is inherently boolean, such as the digital read, always use :

if (thingToTest)

You simply don't need the == operator at all.

@fastrunner08

Posted by: @will

You can't assign a new value to a literal because "1" already has a distinct literal value that can't be modified. 

Note that the compiler should also prevent you from assigning to a const value so:

const  int  myConst = 5;

if (const=6)

should also result in a nastygram from the compiler.

Sorry for the late reply, was late and had to get some sleep.

@will has provided an excellent explanation - I'd just add that, a literal (if it wasn't already clear), is actually a constant!  For example, 1 = 42? huh? How can we assign 42 to 1? We can't, because 1 is not a variable - it is actually a fixed constant set at compile time.

HTH,

Cheers