Gordon Henderson wrote:
> That's not what that statement is doing.
>
> One = is an assignment, 2 == is comparision.
>
> So
> if (a = a + 1024)
>
> evaluates a + 1024, then assigns the result to a, then tests the result
> (of the new a) for true or false ... (or zero or non zero) so it'll always
> be true...
Ok, That makes alot of sense, as i was destroying "a" by assigning to it.
i probably used this in basic51. and got confused.
>
>
> Drop this,
>
>> if(a = a + 1024)
>
> replace with:
>
> if ((a % 1024) == 0)
>
Yes, thank you That is what i wanted it to do.
>> {
>> !P3_0; // if true every K toggles led
>> }
>> *abs_ptr = 0x69 ;
>> *abs_ptr++ ;
>> P1 = a; // status of "a" variable on port P1
>> }
>> _asm ljmp 0 _endasm; // fin.
>> }
>
> The % operator is modulo and an integer division and finds the remainder.
> If this isn't acceptable, then keep a separate variable and increment it
> and compare to 1024. (or initialise it to 1024, decrement it and test for
> zero which may be more efficient if you'rel looking for every last ounce
> of speed)
>
> I don't know anything about the 8051 though, but lots about C... Make sure
> 'a' is a variable of enough bits to hold an address (maybe at least 16
> bits if you use things like uint16_t rather than just 'int', etc.
>
> I'm also not sure what
>
> !P3_0;
>
> is doing. I'm not convinced it's actually having any effect, but maybe it
> is due to the way the C compiler is working? (I'd check the assembly) To
> toggle a bit, assuming P3_0 is a bitfield in a structure pointing to some
> hardware, I'd look to using something like:
>
> P3_0 ^= 1 ;
>
> Which is shorthand for
>
> P3_0 = P3_0 ^ 1 ;
>
> the ^ operator is bitwise-xor.
Oops, i meant to put P3_0=!P3_0;
But I like your version of it better.
>
>
> Also...
>
> P1 = a ;
>
> I'd guess that P1 is an 8-bit port, that will get the bottom 8-bits of a -
> but maybe that's the intention?
Yes i just wanted to look at the value of "a" in the simulator.
Well thanks again Gordon, problem solved.
>
> Gordon
>
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