8 Optimisation. 8.2 Machine-Independent Optimisation

Transcription

1 8 8.2 Machine-Independent Eliminating Redundant Instructions The aim of this optimisation is to identify instructions that do not need to be performed, and eliminating them. This can be done at the level of intermediate code or on the assembler code. We will here focus on the intermediate code. 2 1

2 Avoiding recalculation of values: example Look at the code below: int a, b, c, d; a = a+b*c; d = a+b*c; b = a+b*c; Note that b*c will produce the same result each time, as neither b nor c changes. However, the second line will not produce the same result as the first, since the value of a changes between each calculation. However, the third line DOES repeat the same calculation as in the second, since none of the variables in the expression change. 3 Avoiding recalculation of values: example The source code might produce intermediate code like the following: a = a+b*c d = a+b*c b = a+b*c (*, b, c, t1) (+, a, t1, t2) (=, t2,, a) (*, b, c, t3) (+, a,t3, t4) (=, t4,, d) (*, b, c, t5) (+, a, t5, t6) (=, t6,, b) 4 2

3 Avoiding recalculation of values: example We could simplify it as follows: a = a+b*c (*, b, c, t1) (+, a, t1, t2) (=, t2,, a) a = a+b*c (*, b, c, t1) (+, a, t1, t2) (=, t2,, a) d = a+b*c (*, b, c, t3) (+, a, t3, t4) (=, t4,, d) d = a+b*c (+, a, t1, t4) (=, t4,, d) b = a+b*c (*, b, c, t5) (+, a, t5, t6) (=, t6,, b) b = a+b*c (=, t4,, b) 5 Avoiding recalculation of values: Algorithm An algorithm for the previous optimisation is as follows. Again, we only work on sequences of quadruples with a deterministic flow of control (no jumps into the sequence, no conditional jumps) We maintain a table D showing in which quadruple each variable was last changed. The variable depends on that quadruple for its value. To calculate DEP(Opd): if Opd is in D, return its value else return -1 (Opd= operand ) 6 3

4 ALGORITHM A. Initialise: set D to { } B. For each quadruple i: 1. Calculate the most recent quad j on which OPD1 and OPD2 depend (if no dependency, use the first quad). 2. For each quad k from j+1 to i-1, a. test if quad k is equivalent to quad i. A quad is equivalent if the first 3 fields are identical. In the case of :=, the 4 th (RES) field needs to be identical also. b. If equivalent, Delete quad i In the quads after i, change all references to the RES of quad i to the RES of quad k advance i (goto 1) 3. We get here if the current quad i was not deleted: a. If RES is in D, delete it b. push (RES, i) into D 7 Application of Algorithm Let s apply this algorithm to the example we started with: 8 4

5 i = 0 depquad(0) = max( dep(b), dep(c) ) = -1 No quads to match add (t1, 0) to D advance i (t1, 0) 9 (t1, 0) i = 1 depquad(1)= max( dep(a), dep(t1) ) = 0 No quads to match Add (t2, 1) to D advance i 8. (=, t6,, b) 10 5

6 i = 2 depquad(2)= dep(t2) = 1 No quads to match Add (a, 2) to D advance i (a, 2) 11 (a, 2) i = 3 depquad(3)= max( dep(b), dep(c) ) = -1 Match quads 0-2 Quad 0 matches Delete quad 3 Substitute t1 for t3 in later code Advance i (a, 2) 12 6

7 (a, 2) (a, 2), (t4, 4) i = 4 depquad(4)= max( dep(a), dep(t1) ) = 2 Match quads 3-3 No quad matches Add (t4, 4) to D Advance i 13 i = 5 depquad(5)= dep(t4) = 4 No quads to match Add (d, 5) to D Advance i (d, 5) 14 7

8 (d, 5) i = 6 depquad(6)= max(dep(b), dep(c) ) = -1 Match quads 0-5 Quad 0 matches Delete quad 6 Substitute t1 for t5 in later code Advance i (d, 5) 7. (+, a, t1, t6) 15 (d, 5) 7. (+, a, t1, t6) i = 7 depquad(7)= max(dep(a), dep(t1) ) = 2 Match quads 3-5 Quad 4 matches Delete quad 7 Substitute t4 for t6 in later code Advance i (d, 5) 8. (=, t4,, b) 16 8

9 (d, 5) i = 8 depquad(8)= dep(t4), = 4 Match quads 4-7 No match (need RES to match also) Add (b, 8) to D Advance i (d, 5) (b, 8) 8. (=, t4,, b) 8. (=, t4,, b) 17 This result agrees with the result we reached intuitively at the start of the class 8. (=, t4,, b) 18 9

10 8 8.2 Machine-Independent Re-ordering of Operations Reordering Operations Reordering of Operations In many cases, the order of operations in an expression does not affect the result: A + B + float(c) However, the order does affect our ability to recognise repeated calculations The algorithm in the previous section would not catch redundancies in: A = a * (b + c) B = (c + b) * a because the quads produced would not match: (+, b, c, t1) vs (+, c, b, t2) (*, a, t1, t3) vs (*, t2, a, t4) 20 10

11 Reordering Operations Reordering of Operations A solution here is impose order on the operands of a quadruple, where the order is not important (e.g., add, mul, but not in sub, div). Terms in expressions can be reordered as follows : 1. Terms that are not variable nor constant. 2. Indexed Variables (arrays) in alphabetical order. 3. Variables without indexing in alphabetical order. 4. Constants. This will facilitate the location of common subexpressions and the application of other optimizations. 21 Reordering Operations Reordering of Operations: example Assume the following two statements: a=1+c+d+3; b=d+c+2; Applying the re-orderings from the previous slide: a=c+d+1+3; b=c+d+2; and the common subexpression becomes obvious 22 11

12 Reordering Operations Reordering of Operations: example II Reordering does not always allow simplification: a=c+c+d+3; b=d+c+2; Applying the re-orderings from the previous slide: a=c+c+d+1+3; b=c+d+2; The quadruples produced will be as follows: (+, c, c, t1) (+, t1, d, t2) (+, c, d, t3) (+, t3, 2, t4) No redundant quadruples are produced 23 12