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Example : Add the contents of the 2000H: 0500H memory location to contents of 3000H: 0600H and store the result in 5000H: 0700H.
Solution :
Unlike the past example programs, this program mention to the memory locations in different segments, hence, whereas referring to each location, the data segment will have to be initialized newly with the required value. Given figure shows the flow chart. The instruction sequence for the above flow chart is shown along with the comments.
In fact, simply the program performs the addition of two operands which are located in different memory segments. The program has become lengthy only because of data segment register initialization instructions.
Internal Architecture of Microprocessor : The architecture of 8086 provides a number of improvements over 8085 architecture. It supports a, a set of 16-bit registers ,16-bit AL
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