Hi java coder here
im converting a program into delphi for a project
basicaly ive forgotten my whole delphi maths
soo
can anyone convert this into delphi
(l1 & 0x2 ^ 0xffffffff) == -3;
aaa801 0 Newbie Poster
pritaeas 2,194 ¯\_(ツ)_/¯ Moderator Featured Poster
What is ^ for ?
Glenn9999 0 Newbie Poster
It's an equality test.
(11 and $2 xor $FFFFFFFF) = -3
if I understand the syntax right
dausmus 0 Newbie Poster
The question is a bit silly and incomplete (and quite old!), but deserves an answer for those looking for syntactical comparisons:
What the given example does:
&0x2 leaves only bit 1 intact (counting from bit 0)
^ 0xffffffff then sets all the 0 bits to 1's but inverts the results from the above bit 1
Now testing with -3 (which is 0x11111101) then simply says that bit 1 originally had to be a "1" now that it is inverted to "0" to make the comparison true.
So Delphi then is:
(l1 and $2)
and the original java should've been:
(l1 & 0x2)
and eliminate all the other junk.
if the Op just wanted to see the equivalent (silly, broken, and incomplete) syntax in Delphi, it would look like this:
((l1 and $2) xor $ffffffff) = -3; //silly test for bit 1- TRUE if bit 1 is set... the trailing semicolon is out of place for any kind of real statement, but it matches the OP semicolon syntax.
more proper:
if (l1 and $2 <> 0) then
or if just assigning a boolean result from this:
var myBool : boolean;
mybool := (0 <> (l1 and $2)); // unlike C, boolean is not considered a type of integer in Delphi, so we do a logical operation to gain a boolean result
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