How can one instruction give eight or more different branch targets? Use no tables or aligned routines -- I mean to use one instruction to set the flags and then branch with JCC instructions, but you have to explain how it works - the answer is not enough. There are many answers, so don't think it that hard.

{The idea is not to use POPF/LAHF either - that is no fun ;)}

* As a bonus tell us how the most Jcc can be used for the most targets, from the setting of the flags by one instruction?

Example:
``````
bt eax, 1
jc Target_2
Target_1:

...

Target_2:``````
There are only two target from the one BT instruction - how can you use other instructions to set the flags and JCC as many times as you can?
Posted on 2003-07-18 22:49:59 by bitRAKE
``````
shl eax, 1  ;Here is the flag-setting instruction
jp pp

mp:
js mpps

mpms:
jc mpmspc

mpmsmc:
jmp Target1

mpmspc:
jmp Target2

mpps:
jc mppspc

mppsmc:
jmp Target3

mppspc:
jmp Target4

pp:
js ppps

ppms:
jc ppmspc

ppmsmc:
jmp Target5

ppmspc:
jmp Target6

ppps:
jc pppspc

pppsmc:
jmp Target7

pppspc:
jmp Target8
``````

Proof:
0x00000001 shl 1 = Target1
0x80000001 shl 1 = Target2
0x40000001 shl 1 = Target3
0xC0000001 shl 1 = Target4
0x00000000 shl 1 = Target5
0x80000000 shl 1 = Target6
0x40000000 shl 1 = Target7
0xC0000000 shl 1 = Target8

With 3 flags set, you have a 3-bit number with 8 possible values, each corresponding to a different
target. You can't just have jc Target1, jnc Target2 because that gets rid of all of the other possibilites.
You have to put your jumps into a tree form that jumps to a different spot for all 8 values of the 3 flags.

Note: You could get a few extra targets using the zero flag, but not twice as many since i.e the parity flag
will never be set when the zero flag is.
Posted on 2003-07-19 11:51:37 by evwr
evwr, nice work! :)
Posted on 2003-07-19 14:02:27 by bitRAKE