4x3 keypad and an LCD.. need help.

[mackrackit]

If you must have floating points there is a routine over at the MeLabs site.

The easy way for what you are doing is to drop the decimal point.
x = 2 * 1.5

would be
x = 2 * 15 / 10

Then if you are wanting to display a 3.0

LCDOUT, $FE;1,DEC x,".",DEC2 x//100

65535 is the largest value for a WORD size variable. So 125000 over flows, comes back around starting at zero and the result is 59464.

[taisen83]

i tried the codes but the lcd displaying 3.03.

[mackrackit]

OOPPS, I was not thinking straight

What we are doing.
Take the number with one decimal place and multiply by 10 before the PIC sees it.
1.5 is now 15
1.6 is now 16
...

x = 2 * 16
lcdout $FE,1,DEC x / 10,".",DEC x//10

In the above the "x / 10" will return 3.
x//10 will return the remainder 2.

Look in the Pic Basic manual under Math Operators for more fun/confusing stuff

[taisen83]

thanks! that info really helps me..
But i still got another problem. if the equation goes like this x = 26 // 21.. the answer should be 238095238, the decimal part cannot read by the program and the answer is wrong because the limit is 65535 right? how to get the right answer like display 238 only? thanks again..

[mackrackit]

In this case the decimal needs moved three places to the right.
26/21=1.238095...
1000*26/21=1238.095...
x = (1000*26/21)//1000
x=238

You could have

y=26
x = (1000*y/21)//1000

The above will only work if "y" is an integer 22 through 41.
When reading a sensor the range of values are fixed so with a little imagination a constant like in the above case 1000 will work. When another application comes along with a different set of values a different constant and equation will be required.

What is your application besides a keypad?

[Nicmus]

You could have

y=26
x = (1000*y/21)//1000

The above will only work if "y" is an integer 22 through 41.

I'm not sure why any value 0 - 65 will not satisfy your equation. The only requirement to keep it "legal" is to keep 1000*y under 65535. Also if you can afford to loose one digit after the decimal point (0.00x) then the range for y will be 0 – 650.
I guess knowing more about the application will help make the best decision.

Regards,

Nick

[mackrackit]

I'm not sure why any value 0 - 65 will not satisfy your equation. The only requirement to keep it "legal" is to keep 1000*y under 65535. Also if you can afford to loose one digit after the decimal point (0.00x) then the range for y will be 0 – 650.
I guess knowing more about the application will help make the best decision.

Regards,

Nick

I should have said 2 - 41 .
Beyond these values you run into the leading zero problem with the equation I gave.
Some values above 41 will work, some will return 0.0xx. The 0 tenths is the problem.
Thanks for catching that.