1 reply to this topic

[darknoobie]

Hello forum,

Ive been looking into getting back into the pic micro scene and have been a while since i have programmed some. I pulled up an old delay routine i used before. The $ symbol means the current address right? So if I have :

goto $+1
goto $+1
goto $+1

Since goto requires to cycles.. goto current address 1 time then .... continues for the next 2?
Total cycles used = 6


decfsz timer1, 1 ;Loaded with 0xff
goto $+2

Total number cycles = ?

kinda stuck on last one

[dbug]

As you say $ symbol represents the current address of the instruction containing it, but $+1 has nothing to do with cycles or time. It's simply a way to simplify assembly writing, avoiding having to define too many labels for trivial jumps.

In this case, goto $+1 means: jump to the instruction located at the current address + 1 (only simple arithmetic with addresses). This is the next instruction.

The code you posted is equivalent to this one:

Delay8:   goto Delay6

Delay6:   goto Delay4

Delay4:   goto Delay2

Delay2:

As you can see, unless you need to make calls to these addressess from other locations, it's simpler to write your code.

The total number of cycles needed to execute the 3 jumps are: 2 cycles per jump instruction * 3 jumps = 6 cycles.

I think the last fragment of code you postes is not correct. As it is, without anything else after it, it makes no sense. I think the correct code (for a delay loop) should be:

        decfsz timer1, 1  ; 1 cycle if timer1-1 != 0, 2 cycles if timer-1 == 0

        goto   $-1        ; 2 cycles

This way the loop repeats as many times as timer1 is initialized with. If we start with timer1 = 1, then the first execution takes 2 cycles (because timer1-1 is 0), and the next instruction is not executed (decfsz skips the next instruction if timer1-1 is 0). So it takes 2 cycles when timer1 = 1. If timer1 is greater than 1, then for each value > 1, it will take 1 cycle for decfsz, and 2 cycles for goto. This means 3 cycles * (timer1 - 1).

So the total number of cycles is: 3 * (timer1 - 1) + 2 = 3 * timer1 - 1