1 votes 1 votes Consider following assembly-language program: 1: MOV R3, R7 2: LOAD R8, (R3) 3: ADD R3, R3, 4 4: LOAD R9, (R3) 5: BNE R8, R9, L3 The sum of WAW, RAW, and WAR dependencies from the given code are_________ CO and Architecture test-series computer-architecture data-hazards + – LRU asked Jan 8, 2022 LRU 475 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes WAW I1 – I3 = R3 RAW I1- I2 = R3 I1-I3= R3 I3-I4=R3 I4-I5=R9 I2-I5= R8 WAR I3-I2 = R3 total = WAW ( 1) + RAW ( 5) + WAR ( 1 ) = 7 in instruction 5 R8 & R9 only read the value if their is any mistake please notify me ganesh gaitonde answered Jan 8, 2022 ganesh gaitonde comment Share Follow See all 0 reply Please log in or register to add a comment.