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Example of Constructing the DAG (1)t 1 := 4 * iStep (1):create node 4 and i 0 Step (2):create node Step (3):attach identifier t 1 (2)t 2 := a[t 1 ]Step.

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Presentation on theme: "Example of Constructing the DAG (1)t 1 := 4 * iStep (1):create node 4 and i 0 Step (2):create node Step (3):attach identifier t 1 (2)t 2 := a[t 1 ]Step."— Presentation transcript:

1 Example of Constructing the DAG (1)t 1 := 4 * iStep (1):create node 4 and i 0 Step (2):create node Step (3):attach identifier t 1 (2)t 2 := a[t 1 ]Step (1):create nodes labeled [], a Step (2):find previously node(t 1 ) Step (3):attach label (3)t 3 := 4 * i Here we determine that: node (4) was created node (i) was created node (*) was created just attach t 3 to *. * * t1t1 i0i0 4 * t 1,t 3 i0i0 4 [] t2t2 a0a0

2 Common Subexpression Elimination Detection: Common subexpressions can be detected by noticing, as a new node m is about to be added, whether there is an existing node n with the same children, in the same order, and with the same operator. if so, n computes the same value as m and may be used in its place. Note: The expressions b+c in (1) and (3) are not confused. Example: (1)a:= b + c (2)b:= a – d (3)c:= b + c (4)d:= a - d a:= b + c b:= a – d c:= b + c d:= b + + - b0b0 c0c0 d0d0 c b,d a

3 Dead Code Elimination Transfomation: if x is dead (i.e., not live) – never subsequently used – after a point where statement x := y + z appears in a basic block. Then, this statement may be safely removed. Method: Delete from the DAG any root (node with no ancestors) that has no live variable. Weaker form: delete (when multiple) dead variable label. Repeat until we remove all nodes (labels) that correspond to dead code. a:= b+c b:=b-d c:=c+d e:=b+c + +-+ + + - b0b0 c0c0 d0d0 c b,d a b is not used at the exit of the basic block Examples: a:=b + c d:=a - d c:=d + c e abc c0c0 d0d0 b0b0

4 Renaming Temporary Variables Normal Form: Each statement that defines a temporary, defines a new temporary. - A basic block can always be transformed into an equivalent block which is in normal form Example: (1) t := b + c(1) u := b + c rename + t b0b0 c0c0 + u b0b0 c0c0 Change (rename) label

5 Interchange of Statements Example: t 1 := b + c t 2 := x + y Observation: We can interchange the statements without affecting the value of the block if and only if neither x nor y is t 1 and neither b nor c is t 2, i.e. we have two trees. Note: Normal-form basic blocks permit all statement interchanges that are possible. + t1t1 b0b0 c0c0 + t2t2 x0x0 y0y0

6 Algebraic Transformations Arithmetic Identities: x + 0 = 0 + x = x x – 0 = x x + 1 = 1 + x = x x / 1 = x - Replace left-hand side with simples right hand side. Associative/Commutative laws x + (y + z) = (x + y) + z x + y = y + x Reduction in strength: x ** 2 = x * x 2.0 * x = x + x x / 2 = x + 0.5 - Replace an expensive operator with a cheaper one. Constant folding 2 * 3.14 = 6.28 -Evaluate constant expression at compile time` Methodology: Before we create a node of the DAG we check whether such a node exists modulo the above identities.

7 Global Data Flow Analysis To do code generation and optimization a compiler needs to collect information about the program as a whole and to distribute this information to each block in the flow-graph. Data flow analysis: the process by which the compiler collects data-flow information. - by setting up and solving systems of equations that relate information at various points in the program. Example (generic): out[S] = gen[S] U (in[S] – kill[S]) Information flowing after the execution of S Generated by execution of S Information flowing at the beginning of S Information killed by the exec. of S Note: Sometimes information flows backwards in[S] = … out[S] …

8 Points and Paths Points between two adjacent statements within a block before the first statement after the last statement Paths Consider all points in all the blocks. A path from p 1 to p n is a sequence of points p 1, …, p n such that for each i [1, n-1]: - pi is the point immediately preceding a statement and p i+1 is the point immediately following that statement in the same block, or - p i is the end of some block and p i+1 is the beginning of a successor block. d1: i:= m-1 d2: j:= n flow-graph B d1d1 d2d2 i:= m-1 j:= n State machine

9 Example of DFA Problem: Reaching Definitions Definition of variable x: is a statement that assigns or may assign a value to x. - unambiguous definitions: assignment to x or read & store a value in x from an I/O device. - ambiguous definitions: may define the value of x: procedure call with x as a parameter or global variable. assignment through a pointer. Definition d reaches a point p if there is a path from the point immediately following d to p such that d is not killed along that path. Killing a definition of a variable x happens if between two points along the path there is a definition of x.

10 Reaching Definition (Cont) Inaccuracies: When defining reaching definition we sometimes allow inaccuracies. Undecidability: to decide in general whether each path in a flow graph can be taken is an undecidable problem. Conservative inaccuracy: an inaccuracy is conservative if it never leads to a change in what the program computes. - it is normally conservative to assume that a definition can reach a point even if it may not. - we allow definitions to pass through ambiguous definitions.

11 DF Analysis of Strongly Structured Programs gen[s] = {d} kill[s] = D a – {d} out[s] = gen[s] (in[s] \ kill[s]) gen[s] = gen[s 2 ] (gen[s1] – kill[s 2 ]) kill[s] = kill[s 2 ] (kill[s1] – gen[s 2 ]) in[s 1 ] = in[s], in[s 2 ] = out[s 1 ] out[s] = out[s 2 ] gen[s] = gen[s 1 ] gen[s 2 ] kill[s] = kill[s 1 ] kill[s 2 ] in[s 1 ] = in[s], in[s 2 ] = in[s] out[s] = out[s 1 ] U out[s 2 ] Sd: a:= b + c S S1S1 S2S2 SS1S1 S2S2 (a) (b) (c)

12 DFA of Strongly Structured Programs (Cont) Syntax directed definitions: the equations are syntax directed. - synthesized attributes: gen, kill, (out – depends on in) - inherited attributes: in ( in backward dir – out) Note: out[s] gen[s] (gen – definitions reaching end of S without following paths outside S) Loops & Fixpoints: Given gen[s], kill[s], in[s] we cannot simply use in[s] = in[s1].I = J O in[s1] = in[s] out[s1]O = G ( I – K) out[s] = out[s1]Take O 0 = out[s1] = gen[s1] (in[s1] – kill[s1])I 1 = J 1, O 1 = G ( J – K) I 2 = J G, O 2 = O 1 gen[s] = gen[s1] kill[s] = kill[s1] in[s1] = in[s] gen[s1] out[s] = out[s1] S (d) S1S1

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