Timer0 is used for freq. generation at port 3.5 (Portf) of my AT89C4051 @ 24MHz. Each time the timer overflows the port is toggled. 1st case in asm: ORG 00BH tim0_int cpl Portf reti 2nd case in C51: void timer0_int (void) interrupt 1 { Portf = ~Portf; } The problem in the second case is that a LJMP is added to the service routine and this consumes extra time. How do I implement this in C without the LJMP. Thanks. Han
How do I implement this in C without the LJMP. If you want tight control, use assembler. That you "should not use assembler" is an urban legend spread by incompetents and cheapskate managers that believe that a tyro should be hired in lieu of a pro. The "nobody will be able to maintain it if you use assembler" is only true if the company insist that "a programmer is a programmer" even if they find one willing to work for peanuts. Erik
more You are evidently using the chip to its full extent, congrats. But when doing so, do not expect the developemnt tools to do it for you. Erik
I found a solution and fixed it like this. //Replacement for timer0 routine wich toggles a port #pragma asm CSEG AT 0bh db 0b2h ; cpl db 0b5h ; P3.5 db 32h ; reti #pragma endasm Han.
//Replacement for timer0 routine wich toggles a port #pragma asm CSEG AT 0bh db 0b2h ; cpl db 0b5h ; P3.5 db 32h ; reti #pragma endasm cleaner and simpler: remove the ISR from your C file link an a51 file with ORG Obh col p3.5 reti end
ORG Obh col p3.5 reti Why can nobody make a keyboard that is fit for thumb typing :) the above should, of course be ORG Obh cpl p3.5 reti
It gets better and better: In the C file: // replacement timer0 interrupt #pragma asm CSEG 0bh cpl P3^5 reti #pragma endasm Tnx, Han
I'm sorry, forgot something... #pragma asm CSEG at 0bh cpl P3^5 reti #pragma endasm Han.