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CS Midterm I study guide

# CS Midterm I study guide - Single Precision k = number of...

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k = number of bits in e = 8 n = number of bits in f = 23 Double Precision k = number of bits in e = 8 n = number of bits in f = 23 Floating Point Operator =- × × V 1s M 2E =- × + ×( - - V 1S 1 f e 2k 1 +1) =- × ×( - - ) V 1S f e 2k 1 Type Sign Exponent Significand Total bits Exponent bias Bits precision Single 1 8 23 32 127 24 Double 1 11 52 64 1023 53 Registers Operand Possibilities: Immediate Values o Represented by \$, Hex, and Decimal Registers o Represented by a % Memory Reference o Absolute—points to a direct location i.e.: 0x100 o Indirect—gives a register that stores a memory location i.e.: %eax o Displacement—points to a location past the memory in the register i,e.: c(%eax); add c to the value in %eax and go there o Indexed—adds values of two registers to go to i.e.: c(%eax, %ecx) o Scaled—multiply last register’s memory location by constant i.e.: (,eax,c) Data Formats Byte—b—8 bits Word—w—16 bits Double word—l—32 bits

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CS Midterm I study guide - Single Precision k = number of...

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