Thread: outportb(); Variables?

I am designing a program to work with a circuit I created. I'm just using a 16-bit base DOS compiler because I don't understand the CreateFile() function yet...here's the code:

Code:

#include <stdio.h>
#include <dos.h>

void main(void)
{
	int a;
	char LPT[]="6F28767C395D5F687F7D";

	printf("\n\n\tDisplay Number (0 to 9): ");
	scanf("%i",&a);

	if ((a>9) || (a<0))
	{
		printf("\n\tInvalid Number!  (0 to 9 Only)");
	}

		
	a*=2;
	outportb(0x378,0xLPT[a],LPT[a+1]);
}

When the outportb() function is run, what will it send to the 0x378 location? Say a=0....Would it send 0x6F to the x378 location? I know the code works with printf...it prints out the right array values...I just don't know if I can use a combination of letters, then a variable like that...?

Last question...would using an array like that work faster than just using a switch statement and making a case for each number?

> void main(void)
Nope, its
int main(void)

> outportb(0x378,0xLPT[a],LPT[a+1]);
Read the manual page - does it take 3 params?

Try
outportb(0x378,LPT[a]);
outportb(0x378,LPT[a+1]);

Could I just do:

Code:

outportb(0x378,0xLPT[a]LPT[a+1]);

? If i don't put 0xLPT[a] and I JUST put LPT[a] will it recognize the character has hexadecimal, and send that, or will it send the ASCII character value?

If you really want to send those bytes as hex, then write that in the first place

char LPT[]="\x6F\x28\x76\x7C\x39\x5D\x5F\x68\x7F\x7D";

And drop the a*=2 bit as well, since you now have real hex bytes coded in one character, not two hex-ascii chars

The result is simply
outportb(0x378,LPT[a]);