One Button program for different uses.

Hi friends,

Being a newbie means I have to be humble and ask for help from time to time.
I'm trying to write code for a two button set up, high/low, which are used to change their values in memory. I press either button 3 seconds to enter setup. Once I'm in set up, the same two buttons are inputs to change their respective values in memory. I press the high button 3 seconds enter setup, then the high button increments the H value, and the low button decrements the H value. Likewise, when I press the low button 3 seconds, I enter low value setup, high button increments the low value while low button decrements the low value. If I hold a button down, I want the value to increase/decrease until I reach the desired value.

(forgive me for not following protocol with brackets, I'm still not too good at it)

Code:

if (button4 == HIGH) { tick1 = tick1++; //wait 3 seconds then enter setup. if(tick1 = 65000){ delay(300); tick1 == 0; //reset timer if(button4 == HIGH){ lset == lset--; //decrement low value, write it in memory EEPROM.write(11, lset); //here I want to loop back to button4 to decrement again as needed. } else if(button3 == HIGH){ lset == lset++; //increment low value and write it down delay(300); EEPROM.write(11, lset); //here I want to loop back to button3 to increment again as needed. } } }

I didn't see a question in there anywhere. Perhaps you want to explain exactly what it is you're having trouble with. Perhaps explaining your magic numbers (65000, 300, 11) would help, particularly, why you chose them and how you know they're correct. The following line of code is wrong:

Code:

lset == lset--;

One = is for assignment, two == is for comparison. You also shouldn't assign the result of a -- operator to the same variable. Change those lines to lset--. That subtracts one and stores the value back in lset, like lset = lset - 1.

Also, as for the "bracketing", I assume you mean indentation. Here's a few things to read:
SourceForge.net: Indentation - cpwiki
Indent style - Wikipedia, the free encyclopedia

I recommend one of the top 3 styles from the Wikipedia article, and I find a 4-space indentation a good compromise between being clearly visible and not taking up too much space on the screen. 2 is the absolute smallest you should use, and 8 is the absolute max if you want something wider.

You've got everything turned around in the first few lines.

Code:

tick1 = tick1++; //wait 3 seconds then enter setup. if(tick1 = 65000){ delay(300); tick1 == 0; //reset timer

Should be

Code:

tick1++; //wait 3 seconds then enter setup. (note: no reassignment here) if(tick1 == 65000){ // == for comparison delay(300); tick1 = 0; //reset timer, = for assignment

Thanks friends. I will correct the code. No ==, lset = lset - 1, and turn around the beginning lines. the 65000 ticks represents millis, so that's about 6.5 secs, too long I admit, I'll change that, the 300 millis is a wait between increments, so that the value doesn't jump too fast. 11 is eeprom location.

Quote:

Originally Posted by RobertD

Thanks friends. I will correct the code. No ==, lset = lset - 1,

Or you can simply do lset-- or lset++ it will decrement and increment in place.

Correcting the indentation.

Code:

if (button4 == HIGH) { tick1++; //wait 3 seconds then enter setup. if(tick1 == 35000) { delay(300); tick1 = 0; //reset timer if(button4 == HIGH) { lset == lset--; //decrement low value, write it in memory EEPROM.write(11, lset); delay(300); //here I want to loop back to button4 to decrement again as needed. } else if(button3 == HIGH) { lset = lset + 1; //increment low value and write it down EEPROM.write(11, lset); delay(300); //here I want to loop back to button3 to increment again as needed. } } }

That's not very corrected indentation. Try something like this (notice how subordinate lines of code are more indented than their superiors):

Code:

if (button4 == HIGH) { tick1++; //wait 3 seconds then enter setup. if(tick1 == 35000) { delay(300); tick1 = 0; //reset timer if(button4 == HIGH) { lset--; //decrement low value, write it in memory EEPROM.write(11, lset); delay(300); //here I want to loop back to button4 to decrement again as needed. } else if(button3 == HIGH) { lset++; //increment low value and write it down EEPROM.write(11, lset); delay(300); //here I want to loop back to button3 to increment again as needed. } } }

Also, make note of the changes I highlighed in magenta. Your decrement was still incorrect, and these forms are more succinct and just as readable as the longer version. And since you made a few comments about needing to loop back, I suggest you read the tutorials we have on loops, and Google for some more. Read any textbooks and class notes you have, and work some examples.

Code:

if (button4 == HIGH) { tick1++; //wait 3 seconds then enter setup. if(tick1 == 35000) { delay(300); tick1 = 0; //reset timer if(button4 == HIGH)

Now that the indentation is clear, what does this mean?
The second if seems to be trivially true, if the first one happened.

Are your buttonx variables directly linked to some external hardware?
Are they declared as being 'volatile'?

I have two buttons on hardware, 3 and 4, and I use them both to enter setup and to enter the high and low value in memory, so if I press button 3 for three seconds, I enter setup, then I press 3 again, and this time it increments the high value in memory, while button 4 decrements the value stored in the high memory. Conversely, if I press 4 for three seconds, I set up the low memory with both buttons. That's why I have button4 HIGH again, this time, in setup situation, where both buttons serve to raise/lower the value in high memory. It's a dual use of the same button, first to enter setup, second to increment the value.

And if I let it sit for 5 seconds without changing values, it exits setup.

PS Will read tutorials.

So the question is still: do the button3/4 values change automatically, or do you have to do a read? If the former, make sure you have things marked as volatile so that the compiler won't just optimize all these checks away.

In either case, I don't really see you checking for "holding the button down" -- right now, if the button is pressed, you wait three seconds, and if the button is down then, you go into set mode. Perhaps you want something like "while (button4 == HIGH)" to do the tick counting, and then at the end if the tick count is high enough, go into set mode. (You've also got another design choice to make: what if I hold the button down for five seconds without letting go? Have I changed the value?)

Indeed a "while button down" count 5 seconds, then enter setup, sounds like the right approach.
Then, when either button is held down, I want a tick up/down every 300ms, or at every press of the button, so I can reach the target by holding the button till near, then ticking one tick at the time, forward or back if I overshoot.

Thanks for your help, I will work on this tomorrow. I have to finish this soon, I have customers waiting. :)

Code:

if (button3 == HIGH) { tick1++; //wait 3 seconds then enter setup. if(tick1 == 35000) { digitalWrite(backLightPin, LOW); //flash screen two times. delay(300); digitalWrite(backLightPin, HIGH); delay(300); digitalWrite(backLightPin, LOW); delay(300); digitalWrite(backLightPin, HIGH); delay(300); tick1 = 0; //reset timer hset = EEPROM.read(10); //read and display last setup value lcd.print(hset, DEC); while(button4 == HIGH) { delay(300); hset--; //decrement high value, write. EEPROM.write(11, hset); lcd.print(hset, DEC); } while(button3 == HIGH) { delay(300); hset++; //increment high value, write. EEPROM.write(11, hset); lcd.print(hset, DEC); } //exit setup } } }