[TUHS] C <-> ASM problem

Hi, while creating a web-page about the open questions of how to create C code which compiles through the optimizer run to the same ASM code as the original object was made from, I found the fix for one of my two "top" questions. The (right now) remaining question is here: http://pofo.de/P8000/problems.php The other (solved) problem was: I had the following ASM code: ldk r2,#0 ldb rl2,_u+1060 ld r3,r2 neg r3 add r3,#256 ldb rh3,rl3 clrb rl3 ld _u+48,r3 so i created the following C code out of it: u.u_count = (-u.u_segmts[NUSEGS-1].sg_limit+0x100)<<8; but this compiled to this ASM code: ldk r2,#0 ldb rl2,_u+1060 neg r2 add r2,#256 ldb rh2,rl2 clrb rl2 ld _u+48,r2 As you can see the copy of r2 to r3 and the further processing with r3 is missing here. I also thought "who the fuck would write such an C-code, the code must look different". but I did not found the solution what could have been written in the C code until I've talked today with a colleague of mine at work about ASM and my problems. He isn't familar with Z80(00)-ASM but he used to program ASM years ago with his C64. We took the ASM code and simulated it with values: _u+1060 contains 15 and is loaded to rl2 15 (0x000F / 00000000 00001111) this gets negated (2 complement formed) -15 (0xFFF1 / 11111111 11110001) to that, 256 gets added 241 (0x00F1 / 00000000 11110001) this is 8 bit rightshiftet 61696 (0xF100 / 11110001 00000000) the result gets loaded into _u+48 He then got the idea that all this could be aritmetical written as ((256 - 15)*256) because -15+256 is == 256-15 and rightshifts are done aritmetically by multiplying the value with 256. It could have also been done by having 256² - 256*x. This was great. With that information I wrote in C: u.u_count = (256-u.u_segmts[NUSEGS-1].sg_limit)<<8; And this generated the same ASM code as in the original code problem solved :) -- Oliver Lehmann http://www.pofo.de/ http://wishlist.ans-netz.de/