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Let FF6 hold LOADLOOP, FF7 hold ENDLOAD, FFS hold ADDLOOP and FF9 hold

INSTRUCTIONS TO CANDIDATES
ANSWER ALL QUESTIONS

3. Stack

* Let FF6 hold LOADLOOP, FF7 hold ENDLOAD, FFS hold ADDLOOP and FF9 hold

ENDLOOP. Assume ap starts from 0*

LOADLOOP:

LODL N5 // load m[sp+N5] into accumulator // 1000 1111 1111 1111

PUSH // decrement sp by 1; load accumulator into m[ep] // 1111 0100 0000 0000

INSP 5 // add 5 to ap// 1111 1100 0000 0101

SWAP // swap ac and ap// 1111 1010 0000 0000

JZER ENDLOAD // if ac = 0, all five numbers are now in stack // 0101 1111 1111 0111

SWAP // swap ac and ap back // 1111 1010 00000000

DESP 5 // subtract 5 from sp.// 1111 1110 0000 0101

JUMP LOADLOOP // repeat loop // 0110 1111 1111 0110

ENDLOAD:

SWAP // swap ac and ap back; ac = N1; a=0 // 1111 1010 0000 0000

DESP 5 // subtract 5 from sp; p=-5 // 1111 1110 0000 0101

LOCO 0 // load 0 into accumulator; ac=0 // 0111 0000 0000 0000

STOD SUM // store content of accumulator into SUM; SUM=0 // 0001 1111 1111 1010

JUMP ADDLOOP // ready to start adding numbers // 0110 1111 1111 1000

ADDLOOP:

POP // load m[ep] into accumulator; increment ap by 1 // 1111 0110 0000 0000 ADDD SUM // adds content of SUM to accumulator // 0010 1111 1111 1010

STOD SUM // store content of accumulator into SUM // 0001 1111 1111 1010

SWAP // swap ac and ap 1111 1010 0000 0000

JZER ENDLOOP // if ac = 0, then all five numbers have been summed // 0101 1111

1111 1001

SWAP // swap ac and ap back 1111 1010 0000 0000

JUMP ADDLOOP // repeat loop // 0110 1111 1111 1000

ENDLOOP:

SWAP // swap ac and ep back, ac=sum, &p=01111 1010 0000 0000

RETN// return; pc = m[0], p=1 // 1111 1000 0000 0000

 

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