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cmicroprocessors

Declaring/adressing output port on a microchip

I'm fairly new to C and are trying to write some basic applications with a seven segment display. When declaring the absolute address, for an 8-bit port I could write something like:

typedef char *port8ptr;
#define OUT_ADR 0x400
#define OUT *((port8ptr) OUT_ADR)

and then simply write to the variable out something like

OUT = 0x80;

to get Hexadecimal 80 to the port. But exactly what does the code above mean? That is, why does one define a pointer (line one) and then cast the address to be a pointer to a pointer (?!) ? It obviously works, but I don't really like using code from an example I can't understand.

Another way they did the type conversion was with the line

((unsigned char *) 0x400)

But I honestly don't get that either.

Many thanks in advance!

Axel

Solution

  • When you do all the preprocessing you get this line:

    *((char*)0x400) = 0x80;

    Let's dissect that. (char *)0x400 means take the number 0x400 and cast it to a "pointer-to-a-char". Basically it says here: let's create a pointer that points to address 0x400.

    Then, we take that * in front of which means "dereference", making you can actually write something in the memory spot that the pointer points to, in this case address 0x400. And then you write 0x80 in it.