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CompilersModern Compiler Design

Supplementary Note 1Code Generation (ASU88: Simplified Introduction)

NCYU C. H. Wang

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The Target Machine

The target machine is a byte-addressable machine with four bytes to a word and n general purpose registers.

It has two address instruction of the form: op source, destination

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Op-codes

MOV (move source to destination) ADD (add source to destination) SUB (subtract source from destination)

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Addressing Mode

Literal #c constant c 1

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Examples MOV R0, M

Store the contents of register R0 into memory location M.

MOV 4(R0), M Store the value contents(4+contents(R0)) into memory

location M. MOV *4(R0), M

Store the value contents(contents(4+contrnts(R0))) in to memory location M.

MOV #1, R0 Load the constant 1 into register R0

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Instruction Costs

The cost of an instruction to be one plus the costs associated with the source and destination address modes.

MOV R0, R1 (1) MOV R0, M (2) MOV #1, R0 (2) SUB 4 (R0), *12 (R1) (3)

contents(contents(12+contents(R1)))- contents(4+contents(R0))

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Example

a := b + c1. MOV b, R0 2. MOV b, a ADD c, R0 ADD c, a MOV R0, a (cost=6) (cost=6)

3. R0, R1, R2 contain 4. R1, R2 contain the addresses of a, b and c the values of b and c MOV *R1, *R0 ADD R2, R1 ADD *R2, *R0 MOV R1, a (cost=2) (cost=3)

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Basic Blocks and Flow Graphs

A basic block is a sequence of consecutive statements in which control enters at the beginning and leaves at the end without halt or possibility of branching except at the end.

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Partition into Basic Blocks

First determine the set of leaders The first statement is a leader. Any statement that is the target of a conditional or

unconditional goto is a leader. Any statement that immediately follows a goto or

conditional goto statement is a leader.

For each leader, its basic block consists of the leader and all statements up to but not including the next leader or the end of the program.

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Example (1/2)

Source code

begin prod:=0; i=1; do begin prod :=prod + a[i] * b[i]; i:=i+1 end while i<=20end

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Example (2/2)

Three-address code (the partition)

B1

B2

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Flow Graphs

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A Simple Code Generation (1)

For each three-address statement of the form x:=y op z we perform Invoke a function getreg to determine the

location L where the result of the computation y op z should be stored.

Consult the address descriptor for y to determine y’, the current location y. If the value of y is not already in L, generate the instruction MOV y’ L to place a copy of y in L.

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A Simple Code Generation (2)

Generate the instruction OP z’, L where z’ is a current location of z. Update the address descriptor of x to indicate that x is in location L. If L is a register, update its descriptor to indicate that it contains the value of x, and remove x from all other register descriptor.

If the current values of y and/or z have no next uses, are not live on exit from the block, and are in registers, alter the register descriptor to indicate that, after execution of x:=y op z, those registers no longer will contain y and/or z, respectively.

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Example

d:=(a-b)+(a-c)+(a-c) Three-address code

t:=a-b u:=a-c v:=t+u d:=v+u

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Code sequence

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Generating code for indexed assignments

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Generating code for pointer assignments

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Conditional Statements

CMP x,y CJ< z

Jump to z if the condition code is negative or zero x:=y+z if x<0 goto z

MOV y, R0 ADD z, R0 MOV R0, x CJ< z

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The DAG Representation of Basic Blocks

DAG (Directed Acyclic Graphs) A DAG for a basic block

Leaves labeled by unique identifiers.

Interior nodes labeled by operator symbols

Interior nodes optionally given a sequence of identifiers.

The Interior nodes represent computed values, and the

identifiers labeling a node are deemed to have that

value.

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Example (1/2)

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Example (2/2)

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Constructing a DAG (1/2)

Suppose the current three-address statement is either (i) x:=y op z, (ii) x:=op y or (iii) x:=y. We treat a relational operator like if i<=20 goto as case (i), with x undefined. If node(y) is undefined, create a leaf labeled y,

and let node(y) be this node. In case (i), if node(z) is undefined, create a leaf labeled z and let that leaf be node(z).

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Constructing a DAG (2/2)

In case (i), determine if there is a node labeled op, whose left child is node(y) and whose right child is node(z). If not, create such node. In either event, let n be the node found or created. In case (ii), determine if there is a node labeled op, whose lone child is node(y). If not, create such node, and let n be the node found or created. In case (iii), let n be the node(y).

Delete x from the list of attached identifiers for node(x). Append x to the list of attached identifiers for the node n found in (2) and set node(x) to n.

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Example

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Generating Code from DAGS

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Code sequence

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Rearrange the Order of the Statements

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Other Topics

Code Optimization Dynamic Programming Code Generation

Algorithm