From: BlackJack/Civitas (blackjack_at_civitas64.de)
Date: 2001-02-26 19:36:03
Hi troy,
>[code]
> if (more==1)
> {
> *clearkeybuffer=0;
> VIC.spr_ena=0;
> return pmenu; // *** this is when I want to exit
> }
> *time1=0; // delay the user imput to stop sprite run-a-way
> while (*time1<10)
> {}
> choiceA(pmenu);
This is not a 'jmp choiceA(pmenu)' but a 'jsr'! So it doesn't jump to the
start of this routine but returns if the 'return pmenu' is executed in the
if(more==1) branch. The result is simply thrown away and then the next
line (return 0;) is reached!
A quick fix would be to write 'return choiceA(pmenu);' (and forget the
'return 0;') but then you are using a recursive function call[1] where it
is absolutly unecessary.
The standard way of executing a codeblock again and again is to put it
into an endless loop with 'for (;;) { ...code... }'
or 'while (1) { ...code... }'
Somewhere in ...code... you have to use 'return' or 'break' to exit this
loop.
> return 0; // **** bogus return to make the compiler happy
> // and the reason I keep getting 0 for a carry value
> // what ever I put here get returned. not pmenu
>above.
As I mentioned above this return is not that bogus. If you don't hit the
return key the first time this routine runs, this line will be executed.
CC65 let you leave this return out but inserts it silently for you.
[1] recursive functions call:
Just a quick description. Everytime a function is called it allocates
memory for the given argument(s) on the stack (and this is also true for
local variables that are not declared 'static').
So if you call the same function within the function body, the local
variables are allocated again and are independent from the first call.
Let's have a look at this code:
1:void f(unsigned char x) {
2: if (x>0) f(x-1);
3: printf("%d ", x);
4:}
This will print all numbers from 0 to x. Let's trace 'f(2)' step by step:
Step (Linenumber)
1. (1) x is created on top of the stack and set to 2 (Stack = 2)
2. (2) x is greater than 0 so call f with x-1 = 1 as argument. This is a
jsr not a jmp so we will return here!)
3. (1) x is created on top of the stack and set to 2 (Stack = 1,2)
4. (2) x is still greater than 0 so we call f again with x-1 = 0
5. (1) again a local x is created on the stack (Stack = 0,1,2)
6. (2) now the if is skipped.
7. (3) print x (=0), remove x from the stack (Stack = 1,2) and return
to the point where the function was called (that was in step 4)
8. (3) we are now in line 3 again (right behind the f(x-1);) and print
x (=1), remove x from the stack (Stack = 2) and again return to
the caller (this one was in step 2)
9. (3) print x (=2), remove x from stack and return to the caller.
One popular examples of recursive functions are fibonacci numbers. The
definition is: Every fibonacci number is the sum of the two previous. With
fib(0) = 0 and fib(1) = 1.
unsigned long fib(unsigned long n) {
if (n<2) return n;
return fib(n-1) + fib(n-2);
}
Ciao,
Marc 'BlackJack' Rintsch
--
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This archive was generated by hypermail 2.1.3 : 2001-12-14 22:05:39 CET