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Explain segmentation?
Segment memory addressing divides memory into many segments. Each of these segments can be considered as a linear memory space. Every one of these segment is addressed by a segment register.
Though since the segment register is 16 bit wide and memory needs 20 bits for an address the 8086 appends four bits segment register to attain the segment address. Hence to address the segment 10000H by, say SS register, the SS should contain 1000H.
How many two input AND gates and two input OR gates are required to realize Y = BD+CE+AB ? Ans. Here three product terms, therefore three AND gates of two inputs are needed.
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Q. Drawback of indirect addressing? • Drawback of this scheme is that it needs two memory references to fetch actual operand. First memory reference is to fetch the actual addr
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