Variables of 4

[+atari2600land]

Is there an easy way to check if a variable is a variable of 4 (i.e. if a=4, if a=8, if a=12, etc...)

[Nukey Shay]

If (variable&%00000011) = 0 is true. If either bits 0 or 1 are set, it's not a multiple of 4. You'd need to check if the variable = 0 directly if you wanted to skip that condition.

 

In inline assembly, this is accomplished by

LDA variable

BEQ Is_Zero ;to skip the =zero condition

AND #%00000011 ;throw away the 6 upper bits

BEQ Is_Multiple_Of_Four ;if true

or

BNE Is_Not_Multiple_Of_Four ;if false

Edited August 12, 2010 by Nukey Shay

[+atari2600land]

It doesn't seem to want to work.

snake.bas

[Nukey Shay]

How about if (variable&&%00000011)=0?

(note the double ampersand)

 

Either & or && should correspond to the AND opcode.

[+atari2600land]

Nope, still won't work.

[jbs30000]

I haven't done this, but it should work.

var2=var1//4

var1 is the variable you want to check. //4 means put the remainder of var1/4 in temp1.

If temp1=0 then...

If no remainder then 4 goes into var1 evenly.

[+Random Terrain]

Instead of trying to use this:

 

  if (a && %00000011)=0 && u{0} && v=1 then x=2
  if (e && %00000011)=0 && u{0} && v=1 then x=2
  if (i && %00000011)=0 && u{0} && v=1 then x=2
  if (m && %00000011)=0 && u{0} && v=1 then x=2

 

 

 

What about using this instead:

 

  temp5 = (a & %00000011) : if !temp5 && u{0} && v=1 then x=2
  temp5 = (e & %00000011)  : if !temp5 && u{0} && v=1 then x=2
  temp5 = (i & %00000011)  : if !temp5 && u{0} && v=1 then x=2
  temp5 = (m & %00000011)  : if !temp5 && u{0} && v=1 then x=2

Edited August 12, 2010 by Random Terrain

[Nukey Shay]

Hmm...the use of multiple conditions on the same line might be the problem? Maybe try enclosing the entire expression between parenthesis?

 

ex:

if ((a&%00000011)=0) && u{0} && v=1 then x=2

[+Random Terrain]

Hmm...the use of multiple conditions on the same line might be the problem? Maybe try enclosing the entire expression between parenthesis?

 

ex:

if ((a&%00000011)=0) && u{0} && v=1 then x=2

That doesn't want to compile either (using either type of AND).

[Nukey Shay]

if (a&%00000011)<1 && u{0} && v=1 then x=2

 

This is getting nuts.

[+Random Terrain]

if (a&%00000011)<1 && u{0} && v=1 then x=2

 

This is getting nuts.

That doesn't seem to work either, but both of these compile:

 

temp5 = (a & %00000011) : if temp5=0 && u{0} && v=1 then x=2

temp5 = (e & %00000011) : if temp5=0 && u{0} && v=1 then x=2

temp5 = (i & %00000011) : if temp5=0 && u{0} && v=1 then x=2

temp5 = (m & %00000011) : if temp5=0 && u{0} && v=1 then x=2

 

 

 

 

temp5 = (a & %00000011) : if !temp5 && u{0} && v=1 then x=2

temp5 = (e & %00000011) : if !temp5 && u{0} && v=1 then x=2

temp5 = (i & %00000011) : if !temp5 && u{0} && v=1 then x=2

temp5 = (m & %00000011) : if !temp5 && u{0} && v=1 then x=2

Edited August 12, 2010 by Random Terrain

[Nukey Shay]

Hang on a second...you are constantly rechecking the same conditions. Remove them from the sequence...

 

 

  if !u{0} then skip_the_rest
 if v<>1 then skip_the_rest
 if (a&3)=0 then set_x
 if (e&3)=0 then set_x
 if (i&3)=0 then set_x
 if (m&3)<>0 then skip_the_rest
set_x:
 x=2
skip_the_rest:

Edited August 12, 2010 by Nukey Shay

[+Random Terrain]

Hang on a second...you are constantly rechecking the same conditions. Remove them from the sequence...

Yeah, I just posted his code as-is. Didn't try to clean it up.

[GroovyBee]

I may be wrong but I think bB only does logical AND and not bitwise AND. I don't think the current version makes a distinction between & and &&.

[Nukey Shay]

There should be both (unless something recently broke in it).

[GroovyBee]

My mistake! I did find this from a few years ago :-

 

http://www.atariage.com/forums/topic/100416-conditional-statements-with-score/page__view__findpost__p__1218564

 

By SeaGtGruff saying that bitwise operators don't work in "if" statements. However I don't know if that behaviour has been changed.

[+batari]

My mistake! I did find this from a few years ago :-

 

http://www.atariage.com/forums/topic/100416-conditional-statements-with-score/page__view__findpost__p__1218564

 

By SeaGtGruff saying that bitwise operators don't work in "if" statements. However I don't know if that behaviour has been changed.

If you use the latest build, bitwise AND is supported in "if" statements with some restrictions. For example, you can do "if a & 3 then" but cannot check it against a value. To invert the condition, you need to use "else."

 

This should work:

 if !u{0} then skip_the_rest
 if v<>1 then skip_the_rest
 if a&3 then skip_the_rest else set_x
 if e&3 then skip_the_rest else set_x
 if i&3 then skip_the_rest else set_x
 if m&3 then skip_the_rest
set_x
 x=2
skip_the_rest

[Nukey Shay]

Can you !a&3 That would avoid adding the extra "else" code.

[+batari]

Can you !a&3 That would avoid adding the extra "else" code.

Not currently. bB can evaluate a single bit or a condition between two simple operands and can't currently do anything with those operands (like invert them as you suggest.) Someday I may allow full expressions within conditions as this will vastly improve the language.

 

Anyway, the code may be rewritten as follows, I think:

 if !u{0} then skip_the_rest
 if v<>1 then skip_the_rest
 if a&3 || e&3 || i&3 || m&3 then skip_the_rest
 x=2
skip_the_rest

However, I am not sure if this will save any space over the above code.