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Q Why we use addressing schemes?
An operation code of an instruction tells the operation to be performed. This operation is executed on some data stored in memory or register. Operands can be specified in one of the 3 basic forms which are register, immediate and memory.
Though, why addressing schemes? The question of addressing is connected with how operands are interpreted. We can say in other wordsthat the term "addressing schemes" refers to techniqueused for specifying operands. There are a multitude of instruction formats and addressing schemes. Selecting which schemes are available will affect not only the ease to write compiler but will also consider how efficient the architecture can be?
All computers use more than one addressing schemes to provide programming flexibility to user by offering facilities likeloop control,pointers to memory, indexing of data and program relocation as well as to decrease the number of bits in operand field of instruction. Providing a variety of addressing modes can help decrease instruction counts though having more modes also raises the complexity of machine and in turn can raise the average Cycles per Instruction (CPI). Before we discuss the addressing modes let's consider the notations being used in this section. In description which follows the symbols A, A1, A2 ...... etc. indicate the content of an operand field. So Ai may refer to a memory address or data. In case operand field is a register address then symbols R, R1, R2,... etc., are used. If C indicates the contents (either of an operand field or a register or of a memory location) then (C) indicates the content of the memory location whose address is C.
Q. High Level Language Program Characteristics? So it is clear that new architectures must support high-level language programming. A high-level language system can be implemen
if the 6 is a and the b is 3 what is the c ?
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Prove the following Boolean identities using the laws of Boolean algebra (A + B)(A + C) = A + BC Ans. (A+B)(A+C)=A+BC LHS AA+AC+AB+BC=A+AC+AB+BC OR A((C+1)+A(B+1))+BC
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