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    Comparing Address Machines

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    Compare zero-, one-, two-, and three-address machines by writing programs to compute
    X = (A+B X C)/D - E X F)
    For each of the four machines. The instructions available for use as follows:

    0 Address 1 Address 2 Address 3 Address
    PUSH M LOAD M MOVE (X<-- Y) MOVE (X <--Y)
    POP STORE M ADD (X<--X +Y) ADD (X<-- Y +Z)
    ADD ADD M SUB (X<-- X -Y) SUB (X<-- Y -Z)
    SUB SUB M MUL (X<-- X x Y) MUL (X<-- Y x Z)
    MUL MUL M DIV (X<-- X /Y) DIV (X<-- Y /Z)
    DIV DIV M

    Legend: <-- arrow
    small x multiply

    © BrainMass Inc. brainmass.com October 9, 2019, 8:21 pm ad1c9bdddf
    https://brainmass.com/computer-science/memory/comparing-address-machines-148183

    Solution Preview

    3 Address mode

    X=(A+B*C)/D-E*F

    MUL T,B,C //T=B*C
    ADD T,T,A //T=A+B*C
    DIV T,T,D //T=(A+B*C)/D
    MUL T1,E,F // T1=E*F
    SUB X,T,T1 //X=T-T1=(A+B*C)/D-E*F

    Here we have taken two temporary registers T,T1 and storing the result in these two temp registers and finally, we have stored the content of registers into X which is the result.
    In 3 address machines, each instruction takes 4 memory access, one access to fetch opcode from memory, 2 for getting the 2 inputs operands and a final to write the result back in memory

    Hence, machine generates a total no of 20 ...

    Solution Summary

    The following posting helps compare address machines.

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