En las siguientes versiones (a partir del 80186/80188) estos clculos estn cableados en la mquina y, por lo tanto, cuesta mucho menos tiempo el realizarlos. If we cared about performance, we'd create a 2-character string and use one int 21h system call to print both digits at once, though. direcciones est programado en microcdigo (dentro de la CROM del sistema de control de la unidad de ejecucin).
We could also hoist the MOV AH,02H, because int 21h/ ah=2 doesn't modify AH. We could have done ADD dx, '00' (and changed the cmp) to operate on both nibbles at once (after isolating them each in DH and DL). INT 21H write char to stdout, return in AL. Running an extra add is cheaper than an extra jmp.) input: DL = 0.15 integer (Often you'd implement nibble -> ASCII with a lookup table, but it only needs one branch, not two. The implementation of ONE_DIGIT also has needlessly-complex branching.
#Instruccion mov del emu8086 code#
To save code size, mov dl, 0Fh / and dl, dh would be only 2 bytes per instruction, instead of 3 for and dh, 0Fh, and it takes the mov off the critical path for out-of-order execution. Ejemplo: mov CX,10 Asignamos al registro contador el valor de 10 para que LOOP genere 10 ciclos. Shr dl, cl extract the high nibble into an 8-bit integerĪnd dh, 0Fh clear the high nibble, keeping only the low nibble La instrucción loop decrementa CX en 1, y transfiere el flujo del programa a la etiqueta dada como operando si CX es diferente a 0. This is a simpler and easier-to-understand implementation, which is also more compact and thus faster and better on a real 8086. Using AND to keep only the low 4 bits would be easier, and much faster on a real 8086 (where each count of a shift took a clock cycle, unlike modern CPUs with barrel-shifter ALUs that can do arbitrary shifts in 1 clock cycle). You need one shift to get the high 4 bits down to the bottom of a register. It's getting the high nibble of DL by left-shifting BX, so the two nibbles are split into BH and BL, but the nibble in BL is in the top 4 bytes. mov si, 3: jmp docheckline: thirdline: mov si, 6: jmp docheckline : docheckline: inc cx: mov bh, 0: mov bl, gamepointer si mov al, ds: bx cmp al, '' je checklineloop: inc si: mov bl, gamepointer si cmp al, ds: bx jne checklineloop : inc si: mov bl, gamepointer si cmp al, ds: bx jne checklineloop: mov winflag, 1.
Your code is massively overcomplicated, no wonder it was confusing.
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