SyntaxBomb - Indie Coders

Not sure were to put this but i want to understand a line of Java explained in BlitzMax.

I have a line of code in java which im trying to understand or need translated into Blitzmax. As i have no idea of java im struggling with learning it. So im asking for someones experience/help with it.

F = (result & 0x10000 ? FLAG_C : 0) | overflowAddTable[lookup >> 4] | ((result >> 8) & (FLAG_3 | FLAG_5 | FLAG_S)) | halfcarryAddTable[lookup & 0x07] | ((result & 0xffff) ? 0 : FLAG_Z);
This bit here (result & 0x10000 ? FLAG_C : 0)   Am i right in understanding this as If (result & $1000) then F=FLAG_C else F=0
also,
(result & 0xffff) ? 0 : FLAG_Z) would be If (result & $FFFF) then F=0 else F=FLAG_Z

I see the use of | ie, oring Bits, so i need something to do with bits of the Flags.

Anyone who knows Java or understands the above care to explain more whats happening?

Im trying to get an Adc16() operation altering the correct flags?

Thankyou in advance Baggey

yes, the conditional (or ternary) operator "x ? y : z" is the same as "if x then y else z".

Regarding "|" you need to ensure that whatever you shift around and combine - has a "length" (amount of bits). Especially if the values are of different types (shifting bits of a byte which you add to an int...).


bye
Ron

Wonderful don't you just love it when you have a programming problem and you cant see the wood through the tree's'
Ive been going around in circles with this. DOUBLE CHECK YOUR BITS

Checking for a Carry in a Z80 machine code instruction can have devastating effects with just 1 bit wrong out of 100,000's of codes

I just realized If (result & $1000) then F=FLAG_C else F=0 Is missing a 0 which isnt testing for an overflow of the carry.

It should be If (result & $10000) then F=FLAG_C else F=0

Baggey

I wouldn't even try to remember the bits for this exact reason 
You have FLAG_C so why not have constants for the bits too?

[pseudo code]
' somewhere you have a list of all flag bits
Const FLAG_C_BIT = $10000


' do a check for a flag
If (result & FLAG_C_BIT) F=FLAG_C.....

Not really.  The results of (result & 0x10000 ? FLAG_C : 0) and ((result & 0xffff) ? 0 : FLAG_Z) are not assigned to F directly, but ORed with the rest of the expression.  You could break down the equation like so:
If (result & 0x10000) Then F = FLAG_C Else F = 0
F = F | overflowAddTable[lookup >> 4] | ((result >> 8 ) & (FLAG_3 | FLAG_5 | FLAG_S)) | halfcarryAddTable[lookup & 0x07]
If Not (result & 0xffff) Then F = F | FLAG_Z
Note the Not in the last line, and no Else.  That is because FLAG_Z is ORed only if none of the lower 16 bits of result are set.  If any are set, then doing an OR with 0 results in no change to F.
Another way to deal with it is to use a bit of math, since True casts to 1 and False casts to 0.
F = (((result & 0x10000) <> 0) * FLAG_C) | overflowAddTable[lookup >> 4] | ((result >> 8 ) & (FLAG_3 | FLAG_5 | FLAG_S)) | halfcarryAddTable[lookup & 0x07] | (((result & 0xffff) = 0) * FLAG_Z)