help coding of PIC16F877A

[sfchew7]

CAN ANYONE teach ME HOW TO SIMPLIFY THIS CODE? it is very messy. How if I have 100 parking lots.
I have no idea at all.
1 = available
0 = unavailable
I know how to declare the array
but I dont know how to insert it.
int8 x;
int8 y;
int8 plav[4];

		 if((input(pin_a0)==0) && (input(pin_a1)==1) && (input(pin_a2)==0)) 
		{
				
				lcd_putc("\f\nPL available:2/3");
				
			
					}	
			
				else if((input(pin_a0)==0) && (input(pin_a1)==1) && (input(pin_a2)==1)) 
		{
				
				lcd_putc("\f\nPL available:1/3");
				
			
					}	
			
				else if((input(pin_a0)==1) && (input(pin_a1)==0) && (input(pin_a2)==0)) 
		{
				
				lcd_putc("\f\nPL available:2/3");
				
			
					}	


				else if((input(pin_a0)==1) && (input(pin_a1)==0) && (input(pin_a2)==1)) 
		{
			
				lcd_putc("\f\nPL available:1/3");
				
			
					}	


				else if((input(pin_a0)==1) && (input(pin_a1)==1) && (input(pin_a2)==0)) 
		{
				
				lcd_putc("\f\nPL available:1/3");
				
			
					}

thanks

 

[engshahrul]

If you have 100 inputs and the input is either 1 or 0, you can adding all. Then you will know how many 1's..

 

[betwixt]

If "pin_a0", "pin_a1" and "pin a2" are bits 0, 1 and 2 of a port, treat them as a 3-bit banary value, lets call it "pins_a",

if((pins_a > 0) && (pins_a < 3) lcd_putc("\f\nPL available:1/3");
else lcd_putc("\f\nPL available:2/3");

Does the same as the code you show but I'm not sure it's what you really want.

Brian.

 

[enjunear]

a0	a1	a2	Output
------------------------
0	0	0	?
0	0	1	?
0	1	0	2/3
0	1	1	1/3
1	0	0	2/3
1	0	1	1/3
1	1	0	1/3
1	1	1	?

I think this is what your code is trying to do.

One way to implement this cleanly would be to use a switch/case statement. Take a0, a1, a2 and combine them into one number. I think your naming convention is backwards of normal (a0 is typically the lowest precision bit, thus representing the 1's position... a1 would represent 2', a2 would represent 4's, etc), so this will look a little different than I've typically seen it.

number = 4*a0 + 2*a1 + a0. This will convert your three bits into a number ranging from 0 to 7 (2^3 = 8 binary combinations). Then use code like the following (syntax specific to your device... look in the compiler's help pages for examples).

my_number = 4*a0 + 2*a1 + a0;  //converts three bits to a single unsigned integer value from 0-7

switch(my_number)
{
   case 2, 4:  lcd_print(".....2/3");
   case 3, 5, 6:  lcd_print(".....1/3");
   default;  //default handles all other cases left undefined (e.g. 0, 1 and 7)
}