barry@micomvax.UUCP (02/09/87)
To any experienced Assmebler programmer:
I am working with an IBM AT clone running MS-DOS 3.10 and what i need is the
following:
(1) An assembler routine to do 32-bit by 16-bit division yielding a 32-bit
result (not a 16-bit result as the opcode DIV yields)
Any help anyone out there can give me would be greatly appreciated.
Thanks in advance for all your help.
Please send any and all replies by mail to
...!philabs!micomvax!barry
Barry Kimelman [ UUCP address: ...!philabs!micomvax!barry ]
phone: (514) 744-8200 ext. 2473
Nothing is impossible if you do not have to do it yourself !!
It is not enough that i succeed , everyone else must fail !!herman@ti-csl.UUCP (02/12/87)
in article <766@micomvax.UUCP>, barry@micomvax.UUCP (Barry Kimelman) says: > > I am working with an IBM AT clone running MS-DOS 3.10 and what i need is the > following: > > (1) An assembler routine to do 32-bit by 16-bit division yielding a 32-bit > result (not a 16-bit result as the opcode DIV yields) > > Barry Kimelman [ UUCP address: ...!philabs!micomvax!barry ] > phone: (514) 744-8200 ext. 2473 Simple, use DIV twice, as in: DIV CX ; DX|AX / CX => AX = quotient, DX = remainder XCHG BX,AX ; Exchange quotient with next chunk of dividend DIV CX ; AX = low 16 bits of quotient, DX = remainder This will divide the 48-bit dividend in DX|AX|BX (high|middle|low) by the 16-bit divisor in CX. The 32-bit result will be returned in BX|AX (high|low) with the 16-bit remainder in DX. As long as you stay with unsigned division, this principle can be extended to any length dividend divided by a 16-bit divisor, just by taking the remainder of the previous division as the high 16 bits of the next division, and the next 16 bits of the dividend as the low 16 bits for DIV. -- Herman Schuurman ARPA: herman%TI-CSL@CSNET-RELAY.ARPA Texas Instruments Inc. CSNET: herman@TI-CSL PO Box 226015 M/S 238 USENET: {ut-sally,convex!smu,texsun,rice}!ti-csl!herman Dallas, Texas 75266 VOICE: (214) 995-0845