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
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 Intel 8085 microprocessor uses an 8-bit data bus and a 16-bit address bus. It can address 64 KB of memory, with each memory location capable of storing a byte. Therefore, the maximum positive number that can be represented in an 8-bit register is 255 (2^8 - 1), while in a 16-bit address space, the maximum is 65,535 (2^16 - 1) when considering the entire memory range.
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
The expression 23x 16 is a multiplication operation involving the numbers 23 and 16. When calculated, it equals 368. Thus, the correct classification of 23x 16 is a multiplication expression that results in a product.
As a product of its prime factors: 2*2*2*2 = 16 or as 2^4 = 16
The 8085 was replaced with the 8086/8088. As such, there is no 16 bit version of the 8085.
Because Intel designed it that way. The 8085 was designed as a 8 bit computer in a 16 bit address space. This means that the PC (Program Counter) and SP (Stack Pointer) should be 16 bits in size.
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 has a 16 bit address bus.
The 8085 is an 8-bit microprocessor. Even though there are some 16-bit registers (BC, DE, HL, SP, PC), with some 16-bit operations that can be performed on them, and a 16-bit address bus, the accumulator (A), the arithmetic logic unit (ALU), and the data bus are 8-bits in size, making the 8085 an 8-bit computer.
No. The 8086 has instructions not present in the 8085. The 8086 was marketed as "source compatible" with the 8085, meaning that there was a translator program which could convert assembly language code for the 8085 into assembly language code for the 8086. However, this does not mean that the compiled 8086 assembly code would then run on an 8085; among other things, the 8086 was a true 16-bit processor, as opposed to the 8085 which was an 8-bit processor that supported a few 16-bit operations.
The stack pointer is 16 bits in size on the 8085 because that is how Intel designed it. The address bus is also 16 bits, so it made sense for the program and stack to be located anywhere in that address space.
Because that's how Intel designed it. The 8085 is an 8-bit computer operating on a 16-bit address space.
The 8086 microprocessor includes the instruction for multiplication of unsigned integers, specifically the MUL and IMUL instructions, which are not available in the 8085 microprocessor. While the 8085 has basic arithmetic operations like addition and subtraction, the 8086's support for multiplication and division instructions allows for more complex arithmetic operations directly in hardware. Additionally, the 8086's capability to handle larger data sizes (16-bit) further distinguishes its arithmetic capabilities from the 8-bit 8085.
Because it is an 8-bit microprocessor.
The 8086/8088 microprocessor family is a 16 bit microprocessor. The 8086 implementation also has a 16 bit data bus, but the 8088 implementation has an 8 bit data bus, comparable to the 8085. The 8088 implementation was intended as a logical upgrade from the 8085, while keeping the complexity of the system on an equal footing as the 8085.
16 bit addition