Lxi b, 0000h lhld 8000h xchg lhld 8002h dcx d l006: lda 8002h add l mov l, a lda 8003h adc h mov h, a jnc l013 l013: inx b dcx d mov a, d ora e jnz l006 shld 8006h mov l, c mov h, b shld 8004h hlt
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The most significant difference between the Intel 8085 and 8086 microprocessors is that the 8085 is an 8-bit system and the 8086 is a 16-bit system. This difference allows the 8086 system to have a much larger set of operational instructions and can make calculations to more significant places. Note: the 8085 processor does have two 16-bit registers. The pointer and the program counter.
The 8085 is an 8-bit computer, with only limited capability to do 16 bit arithmetic. In order to add two 16-bit numbers, NUM1 and NUM2, together, and store the result at NUM3, you can use the code... LHLD NUM1 XCHG LHLD NUM2 DAD D SHLD NUM3 If you want to subtract NUM1 from NUM2, you need to take the two's complement first, by inverting it and adding one... LHLD NUM1 MOV A,H CMA MOV H,A MOV A,L CMA MOV L,A INX H ... and then continue with adding NUM2... XCHG LHLD NUM2 DAD D SHLD NUM3
As a product of its prime factors: 2*2*2*2 = 16 or as 2^4 = 16
2 x 2 x 2 x 2 = 16
multiplication is commutative. interpret % as the fraction 1/100. 16 percent of 40 = 16 x 1/100 x 40 40 percent of 16 = 40 x 1/100 x 16 same numbers in any order and you get the same answer when the only operation is multiplication.