Port_C and tris problem with 18f452

[guess]

I have been using PCWH 3.221. I created my project using wizart. I defined the PIN_C7, PIN_C6, PIN_C5 as a input. The other pins as a output. If a use the following statements , The green and red led don't light.

#define Green_Led PIN_C3
#define Red_Led PIN_C2

#use fast_io (C)
output_high(Red_Led);
output_High(Green_Led);

if I deleted the statement "#use fast_io (C)" then led is giving light. But this time the other pin c7 and c6 create the problem.

#include <18F452.h>
#device adc=10
#FUSES NOWDT, WDT128, HS, NOPROTECT, NOOSCSEN, NOBROWNOUT, BORV20, NOPUT, NOSTVREN, NODEBUG, NOLVP, NOWRT, NOWRTD, NOWRTB, NOWRTC, NOCPD, NOCPB, NOEBTR, NOEBTRB

#use delay(clock=20000000)

[newguy]

Don't trust the wizard to properly set things up. If it did function correctly, there should be this in the wizard-generated code:

[Guest2]

Ok I find the problem But what is the reason the following. I didn't understand. (Effect of #use fast_io (C))


#use fast_io (C)
output_high(Red_Led);
output_High(Green_Led);

if I deleted the statement "#use fast_io (C)" then led is giving light.

[Ttelmah]

With 'fast_io' selected, _you_ must set the TRIS variable yourself, and get it right. Otherwise, the compiler will _automatically_ set the TRIS, so a pin is an output, if you perform an output instruction, and so it is an input, if you perform an input instruction. Fast_io is _faster_, and also means that you can (for example), elect to 'setup' a pin,and then change the TRIS to output the data when required. Hence to work with fast_io selected, you need the tris line suggested by newguy, or the pins will remain as inputs (the default on wake up).

Best Wishes

[newguy]

The compiler has been written in such a way as to take care of a lot of "behind the scenes" things automatically for you. One of those things are the port direction registers.

With some exceptions, almost all PIC port pins can be configured as input (1) or output (0). If, in your C code, you set a particular pin high or low, then obviously you want that pin to be an output. The compiler will then add extra instructions, hidden from you, that make the appropriate pin an output first, then actually set the pin high or low. Same thing goes for reading a pin's state: extra instructions are added to make the pin an input before it is read.

This takes place by default, unless you tell the compiler "no, I'll take care of the port direction myself, thank you very much."

To let the compiler take care of things: no extra directives are needed, this is the default state. Can also be invoked by