Compiling Techniques

Transcription

1 Lecture 11: Introduction to 13 November 2015

2 Table of contents 1 Introduction Overview The Backend The Big Picture 2 Code Shape

3 Overview Introduction Overview The Backend The Big Picture Source code FrontEnd Middle End BackEnd Machine Code Errors Front-end Lexer Parser AST builder Semantic Analyser Middle-end Optimizations (Compiler Optimisations course)

4 The Back end Introduction Overview The Backend The Big Picture Instruction Selection Register Allocation Instruction Scheduling Machine code Errors Translate into target machine code Choose instructions to implement each operation Decide which value to keep in registers Ensure conformance with system interfaces Automation has been less successful in the back end

5 Instruction Selection Introduction Overview The Backend The Big Picture Instruction Selection Register Allocation Instruction Scheduling Machine code Errors Mapping the into assembly code (in our case AST to Java ByteCode) Assumes a fixed storage mapping & code shape Combining operations, using address modes

6 Register Allocation Introduction Overview The Backend The Big Picture Instruction Selection Register Allocation Instruction Scheduling Machine code Errors Deciding which value reside in a register Minimise amount of spilling

7 Instruction Scheduling Introduction Overview The Backend The Big Picture Instruction Selection Register Allocation Instruction Scheduling Machine code Errors Avoid hardware stalls and interlocks Reordering operations to hide latencies Use all functional units productively Instruction scheduling is an optimisation Improves quality of the code. Not strictly required.

8 The Big Picture Introduction Overview The Backend The Big Picture How hard are these problems? Instruction selection Can make locally optimal choices, with automated tool Global optimality is NP-Complete Instruction scheduling Single basic block heuristic work quickly General problem, with control flow NP-Complete Register allocation Single basic block, no spilling, 1 register size linear time Whole procedure is NP-Complete (graph colouring algorithm) These three problems are tightly coupled! However, conventional wisdom says we lose little by solving these problems independently.

9 Overview The Backend The Big Picture How to solve these problems? Instruction selection Use some form of pattern matching Assume enough registers or target important values Instruction scheduling Within a block, list scheduling is close to optimal Across blocks, build framework to apply list scheduling Register allocation Start from virtual registers & map enough into k With targeting, focus on good priority heuristic Approximate solutions Will be important to define good metrics for close, good, enough,....

10 Generating Code for Register-Based Machine The key code quality issue is holding values in registers when can a value be safely allocated to a register? When only 1 name can reference its value Pointers, parameters, aggregate & arrays all cause trouble when should a value be allocated to a register? when it is both safe & profitable Encoding this knowledge into the assign a virtual register to anything that go into one load or store the others at each reference Register allocation is key All this relies on a strong register allocator.

11 Register-based machine Most real physical machine are register-based Instruction operates on registers. The number of architecture register available to the compiler can vary from processor to processors. The first phase of code generation usually assumes an unlimited numbers of registers (virtual registers). Later phases (register allocator) converts these virtual register to the finite set of available physical architectural registers (more on this in lecture on register allocation).

12 Generating Code for Register-Based Machine Memory x y Example: x+y l o a d r1 // l o a d the a d d r e s s o f x i n t o r1 loada r1 r2 // now v a l u e o f x i s i n r2 l o a d r3 // l o a d the a d d r e s s o f y i n t o r3 loada r3 r4 // now v a l u e o f y i s i n r4 add r2, r4 r5 // r5 c o n t a i n s x+y

13 Exercise Write down the list of assembly instructions for x+(y*3) Exercise Assuming you have an instruction muli (multiply immediate), rewrite the previous example. This illustrate the instruction selection problem (more on this in following lectures).

14 Visitor Implementation for binary operators Binary operators R e g i s t e r v i s i t B i n O p ( BinOp bo ) { R e g i s t e r l h s R e g = bo. l h s. a c c e p t ( t h i s ) ; R e g i s t e r l h s R e g = bo. r h s. a c c e p t ( t h i s ) ; R e g i s t e r r e s u l t = n e x t R e g i s t e r ( ) ; switch ( bo. op ) { case ADD: emit ( add l h s R e g. i d rhsreg. i d r e s u l t. i d ) ; break ; case MUL: emit ( mul l h s R e g. i d rhsreg. i d r e s u l t. i d ) ; break ;... } r e t u r n r e s u l t ; }

15 Visitor Implementation for variables x l o a d r1 // l o a d the a d d r e s s o f x i n t o r1 loada r1 r2 // now v a l u e o f x i s i n r2 Var R e g i s t e r v i s i t V a r ( Var v ) { R e g i s t e r addrreg = n e x t R e g i s t e r ( ) ; R e g i s t e r r e s u l t = n e x t R e g i s t e r ( ) ; emit ( l o a d I v. a d d r e s s addrreg. i d ) ; emit ( loada addrreg. i d r e s u l t. i d ) ; r e t u r n r e s u l t ; }

16 Visitor Implementation for integer literals 3 l o a d I 3 r1 IntLiteral R e g i s t e r v i s i t I n t L i t e r a l ( I n t L i t e r a l i t ) { R e g i s t e r r e s u l t = n e x t R e g i s t e r ( ) ; emit ( l o a d I i t. v a l u e r e s u l t. i d ) ; r e t u r n r e s u l t ; }

17 Generating Code for Stack-Based Virtual Machine Stack-based machine don t use registers but instead perform computation using an operand stack (e.g., Java Bytecode, see previous lecture). Makes it easier to write a code generator, no need to handle register allocation. Producing the real register-based machine code is delayed to runtime (jit-compiler).

18 Local variables 0 1 x y Example: x+y using Java Bytecode i l o a d 0 // push x onto the s t a c k i l o a d 1 // push y onto the s t a c k i a d d // top o f s t a c k c o n t a i n s x+y

19 Binary operators Operand stack lhs value rhs value bo value... Void v i s i t B i n O p ( BinOp bo ) { bo. l h s. a c c e p t ( t h i s ) ; bo. r h s. a c c e p t ( t h i s ) ; switch ( bo. op ) { case ADD: emit ( i a d d ) ; break ; case MUL: emit ( i m ul ) ; break ;... } r e t u r n n u l l ; }

20 Identifiers and Integer Literals Void v i s i t V a r ( Var v ) { emit ( i l o a d v. l o c a l V a r I d ) ; } Void v i s i t I n t L i t e r a l ( I n t L i t e r a l i t ) { i n t v a l u e = i t. v a l u e ; i f ( i t. v a l u e < 32768) emit ( s i p u s h i t. v a l u e ) ; e l s e... } Exercise Unfortunately, Java ByteCode does not have an instruction for pushing full integer. Need to generate code for that. Complete the code above.

21 Next lecture: Code Shape Conditions Function calls Loops If statement