Maltanto

// IN: (OFFSET_Y, OFFSET_X) // OUT: HL=VRAM_NAME ADDRESS, A=PATTERN LD A,(OFFSET_Y) AND #F8 // V7 V6 V5 V4 V3 0 0 0 LD H,0 LD L,A ADD HL,HL // 0 0 0 0 0 0 0 V7 | V6 V5 V4 V3 0 0 0 0 ADD HL,HL // 0 0 0 0 0 0 V7 V6 | V5 V4 V3 0 0 0 0 0 LD DE,VRAM_NAME ADD HL,DE // HL=VRAM_NAME+(Y*32) LD A,(OFFSET_X) // X7 X6 X5 X4 X3 X2 X1 X0 SRL A SRL A SRL A // 0 0 0 X7 X6 X5 X4 X3 OR L LD L,A // V5 V4 V3 X7 X6 X5 X4 X3 // READ VRAM // THIS IS OK ON EMULATORS, BUT MOST CERTAINLY WON'T WORK ON THE REAL HARDWARE DUE TO TIMING CONSTRAINTS DI LD A,L OUT (#BF),A LD A,H AND #3F OUT (#BF),A IN A,(#BE) EI RET

Just a couple of things, if I may.. I think there's a potential bug lurking right here: SRL A ; divide by 8 (192/8) 24 Rows (actually divide 3 times by 2) SRL A SRL A LD B, A .. <enter loop with B> When OFFSET_Y is in the range of (0..7) the result after dividing by 8 will be 0 and that will cause the loop to repeat 256 times. You should check for Z after the last SRL A and skip the loop if needed. Anyway, I think the approach of dividing by 8 and then multiplying that same value by 32 is a bit cumbersome and unnecessary . Here's an easier alternative to calculate the pattern address for a given Y_OFFSET in A. I've put some comments laying out the bit fields to help visualize what's going on: //A=OFFSET_Y // V7 V6 V5 V4 V3 V2 V1 V0 AND #F8 // V7 V6 V5 V4 V3 0 0 0 LD H,0 LD L,A ADD HL,HL // 0 0 0 0 0 0 0 V7 | V6 V5 V4 V3 0 0 0 0 ADD HL,HL // 0 0 0 0 0 0 V7 V6 | V5 V4 V3 0 0 0 0 0 LD DE,VRAM_NAME ADD HL,DE // HL=VRAM_NAME+(Y*32) I'm taking advantage of the fact that we're multiplying an unknown value (OFFSET_Y) by a power of 2. Also that zeroing the lowest 3 bits of something is in fact the same as dividing by 8 and then multiplying the result again by 8, so we just need to multiply by 4 to get the proper offset (Y*32). This way there's no need for expensive loops nor safety checks when B=0. The OFFSET_X part is even easier. We know the lowest 5 bits of the adddress in HL are always 0, so we can just divide X by 8 and add the result to the L register: LD A,(OFFSET_X) // X7 X6 X5 X4 X3 X2 X1 X0 SRL A SRL A SRL A // 0 0 0 X7 X6 X5 X4 X3 OR L LD L,A // V5 V4 V3 X7 X6 X5 X4 X3 Regards Jorge