passing a port around.

Overclocked said:

I might be a bit late on this but this is how Ive done it: (XC8 Complier)

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#define LED LATAbits.RA5
#define FET LATAbits.RA4
...
LED = 1;
FET = 1;

misterT said:

Accessing a port is simply about accessing the register (memory location) that controls that port. So, all you need to do is "pass around" the memory address of the register.

example:
register address: 0x12

function prototype to accept memory locations:
void writeToMemory(void *address);

pass it to a function (as a pointer) and use it:
(uint8_t*)0x12 //cast memory address to pointer (the compiler now knows that the memory address is for unsigned 8-bit variable at that address)

dereference the pointer so that you can write to that location. Without dereferencing you would just change the address, not the value at that address.
(*((uint8_t*)0x12)) = 0b11111111; // write all ones to the port

If you look at how registers and other ports are defined in your library you will find something like:
#define PORTA (*((uint8_t*)0x12))

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NorthGuy said:

You can use a pointer, because you need to pass a reference to the sfr location. For example:

Code:

typedef char* port;

void setPin (port myPort, int pin)
{
    *myPort |= (1<<pin);
}

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Isn't it a problem to assign int value into Char Pointer?

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misterT said:

*myPort |= (1<<pin);

I think that in this case the compiler does an implicit cast to the correct type.
Notice that the "pin" variable is not the value that is assigned to the port. It is just the number of left shifts applied to the constant 1.

You can do an explicit cast if you want.
*myPort |= ((char)(1<<pin));

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KISS said:

I think an unsigned int would be a better choice, but int would likely work depending on the address space.

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electroRF said:

NorthGuy said:
I think an unsigned int would be a better choice, but int would likely work depending on the address space.

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