in CO and Architecture retagged by
763 views
0 votes
0 votes
$I_1 :  MUL \ \ \ \ \   R_1,  R_2, R_3$        // $R_1 \leftarrow R_2 \times R_3$

$I_2 :  ADD \ \ \ \ \   R_4,  R_4, R_1$

$I_3 :  MUL \ \ \ \ \   R_1,  R_5, R_6$

$I_4 :  SUB \ \ \ \ \   R_4,  R_4, R_1$

 

Sum of RAW, WAR and WAW dependencies is  _____.
in CO and Architecture retagged by
763 views

4 Comments

Is this question is self made?
0
0
partially, i don't have answer key for this
0
0
Answer should be 7

I1 - I4 should not be considered for RAW dependency...
0
0

2 Answers

0 votes
0 votes
Answer should be 7

RAW = 3

WAW = 2

WAW = 2
0 votes
0 votes
3+2+2=7

RAW(1-2,2-4,3-4)

WAR(2-3,2-4)

WAW(1-3,2-4)

Related questions