THEORY OF COMPILATION
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1 Lecture 10 Code Generation THEORY OF COMPILATION EranYahav Reference: Dragon 8. MCD
2 You are here Compiler txt Source Lexical Analysis Syntax Analysis Parsing Semantic Analysis Inter. Rep. (IR) Code Gen. exe Executable text code 2
3 Last Week: Runtime Part II Nested procedures Object layout Inheritance Multiple inheritance 3
4 Today Runtime checks Garbage collection Generating assembly code 4
5 Runtime checks generate code for checking attempted illegal operations Null pointer check MoveField, MoveArray, ArrayLength, VirtualCall Reference arguments to library functions should not be null Array bounds check Array allocation size check Division by zero If check fails jump to error handler code that prints a message and gracefully exists program 5
6 Null pointer check # null pointer check cmp $0,%eax je labelnpe Single generated handler for entire program labelnpe: push $strnpe call println push $1 call exit # error message # error code 6
7 Array bounds check # array bounds check mov -4(%eax),%ebx # ebx = length mov $0,%ecx # ecx = index cmp %ecx,%ebx jle labelabe # ebx <= ecx? cmp $0,%ecx jl labelabe # ecx < 0? Single generated handler for entire program labelabe: push $strabe call println push $1 call exit # error message # error code 7
8 Array allocation size check # array size check cmp $0,%eax # eax == array size jle labelase # eax <= 0? Single generated handler for entire program labelase: push $strase # error message call println push $1 # error code call exit 8
9 Automatic Memory Management automatically free memory when it is no longer needed not limited to OO programs, we show it here because it is prevalent in OO languages such as Java also in functional languages approximate reasoning about object liveness use reachability to approximate liveness assume reachable objects are live non-reachable objects are dead Three classical garbage collection techniques reference counting mark and sweep copying 9
10 GC using Reference Counting add a reference-count field to every object how many references point to it when (rc==0) the object is non reachable non reachable => dead can be collected (deallocated) 10
11 Managing Reference Counts Each object has a reference count o.rc A newly allocated object o gets o.rc = 1 why? write-barrier for reference updates update(x,old,new) { old.rc--; new.rc++; if (old.rc == 0) collect(old); } collect(old) will decrement RC for all children and recursively collect objects whose RC reached 0. 11
12 Cycles! cannot identify non-reachable cycles reference counts for nodes on the cycle will never decrement to 0 several approaches for dealing with cycles ignore periodically invoke a tracing algorithm to collect cycles specialized algorithms for collecting cycles 12
13 GC Using Mark & Sweep Marking phase mark roots trace all objects transitively reachable from roots mark every traversed object Sweep phase scan all objects in the heap collect all unmarked objects 13
14 GC Using Mark & Sweep mark_sweep() { for Ptr in Roots mark(ptr) sweep() } mark(obj) { if mark_bit(obj) == unmarked { mark_bit(obj)=marked for C in Children(Obj) mark(c) } } Sweep() { } p = Heap_bottom while (p < Heap_top) if (mark_bit(p) == unmarked) then free(p) else mark_bit(p) = unmarked; p=p+size(p) 14
15 Copying GC partition the heap into two parts: old space, new space GC copy all reachable objects from old space to new space swap roles of old/new space 15
16 Example old new A B Roots C D E 16
17 Example old new A B A Roots C C D E 17
18 Summary How objects are organized in memory Automatic management of memory Coming up Generating assembly code 18
19 target languages IR + Symbol Table Code Gen. Absolute machine code Relative machine code Assembly 19
20 From IR to ASM: Challenges mapping IR to ASM operations what instruction(s) should be used to implement an IR operation? how do we translate code sequences call/return of routines managing activation records memory allocation register allocation optimizations 20
21 Intel IA-32 Assembly Going from Assembly to Binary Assembling Linking AT&T syntax vs. Intel syntax We will use AT&T syntax matches GNU assembler (GAS) 21
22 AT&T vs. Intel Syntax Attribute AT&T Intel Parameter order Parameter Size Immediate value signals Effective addresses Source comes before the destination Mnemonics are suffixed with a letter indicating the size of the operands (e.g., "q" for qword, "l" for dword, "w" for word, and "b" for byte) Prefixed with a "$", and registers must be prefixed with a "% General syntax DISP(BASE,INDEX,SCALE) Example: movl mem_location(%ebx,%ecx,4), %eax Destination before Derived from the name of the register that is used The assembler automatically detects the type of symbols; i.e., if they are registers, constants or something else. Use variables, and need to be in square brackets; additionally, size keywords like byte, word, or dword have to be used.[1] Example: mov eax, dword [ebx + ecx*4 + mem_location] 22
23 IA-32 Registers Eight 32-bit general-purpose registers EAX accumulator for operands and result data. Used to return value from function calls. EBX pointer to data. Often use as array-base address ECX counter for string and loop operations EDX I/O pointer (GP for us) ESI GP and source pointer for string operations EDI GP and destination pointer for string operations EBP stack frame (base) pointer ESP stack pointer EFLAGS register EIP (instruction pointer) register Six 16-bit segment registers (ignore the rest for our purposes) 23
24 Not all registers are born equal EAX EDX Required operand of MUL,IMUL,DIV and IDIV instructions Contains the result of these operations Stores remainder of a DIV or IDIV instruction (EAX stores quotient) ESI, EDI ESI required source pointer for string instructions EDI required destination pointer for string instructions Destination Registers of Arithmetic operations EAX, EBX, ECX, EDX EBP stack frame (base) pointer ESP stack pointer 24
25 IA-32 Addressing Modes Machine-instructions take zero or more operands Source operand Immediate Register Memory location (I/O port) Destination operand Register Memory location (I/O port) 25
26 Immediate and Register Operands Immediate Value specified in the instruction itself GAS syntax immediate values preceded by $ add $4, %esp Register Register name is used GAS syntax register names preceded with % mov %esp,%ebp 26
27 Memory and Base Displacement Operands Memory operands Value at given address GAS syntax - parentheses mov (%eax), %eax Base displacement Value at computed address Address computed out of base register, index register, scale factor, displacement offset = base + (index*scale) + displacement Syntax: disp(base,index,scale) movl $42, $2(%eax) movl $42, $1(%eax,%ecx,4) 27
28 Base Displacement Addressing Array Base Reference (%ecx,%ebx,4) Mov (%ecx,%ebx,4), %eax offset = base + (index*scale) + displacement %ecx = base %ebx = 3 offset = base + (3*4) + 0 = base
29 How do we generate the code? break the IR into basic blocks basic block is a sequence of instructions with single entry (to first instruction), no jumps to the middle of the block single exit (last instruction) code execute as a sequence from first instruction to last instruction without any jumps edge from one basic block B1 to another block B2 when the last statement of B1 may jump to B2 29
30 Example B1 t 1 := 4 * i t 2 := a [ t 1 ] if t 2 <= 20 goto B 3 False True B2 t 3 := 4 * i t 4 := b [ t 3 ] goto B4 B3 t 5 := t 2 * t 4 t 6 := prod + t 5 prod := t 6 goto B 4 B4 t 7 := i + 1 i := t 2 Goto B 5 30
31 creating basic blocks Input: A sequence of three-address statements Output: A list of basic blocks with each threeaddress statement in exactly one block Method Determine the set of leaders (first statement of a block) The first statement is a leader Any statement that is the target of a conditional or unconditional jump is a leader Any statement that immediately follows a goto or conditional jump 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 31
32 control flow graph A directed graph G=(V,E) nodes V = basic blocks edges E = control flow (B1,B2) E when control from B1 flows to B2 prod := 0 i := 1 t 1 := 4 * i t 2 := a [ t 1 ] t 3 := 4 * i t 4 := b [ t 3 ] t 5 := t 2 * t 4 t 6 := prod + t 5 prod := t 6 t 7 := i + 1 i := t 7 if i <= 20 goto B 2 B 1 B 2 32
33 example CFG source IR B 1 i = 1 B 2 j = 1 1) i = 1 for i from 1 to 10 do for j from 1 to 10 do a[i, j] = 0.0; for i from 1 to 10 do a[i, i] = 1.0; 2) j =1 3) t1 = 10*I 4) t2 = t1 + j 5) t3 = 8*t2 6) t4 = t3-88 7) a[t4] = 0.0 8) j = j + 1 9) if j <= 10 goto (3) 10) i=i+1 11) if i <= 10 goto (2) B 3 B 4 t1 = 10*I t2 = t1 + j t3 = 8*t2 t4 = t3-88 a[t4] = 0.0 j = j + 1 if j <= 10 goto B3 i=i+1 if i <= 10 goto B2 12) i=1 13) t5=i-1 B 5 i = 1 14) t6=88*t5 15) a[t6]=1.0 16) i=i+1 17) if I <=10 goto (13) B 6 t5=i-1 t6=88*t5 a[t6]=1.0 i=i+1 if I <=10 goto B6 33
34 Variable Liveness A statement x = y + z defines x uses y and z A variable x is live at a program point if its value is used at a later point y = 42 z = 73 x = y + z print(x); x undef, y live, z undef x undef, y live, z live x is live, y dead, z dead x is dead, y dead, z dead (showing state after the statement) 34
35 Computing Liveness Information between basic blocks dataflow analysis (next lecture) within a single basic block? idea use symbol table to record next-use information scan basic block backwards update next-use for each variable 35
36 Computing Liveness Information INPUT: A basic block B of three-address statements. symbol table initially shows all non-temporary variables in B as being live on exit. OUTPUT: At each statement i: x = y + z in B, liveness and next-use information of x, y, and z at i. Start at the last statement in B and scan backwards At each statement i: x = y + z in B, we do the following: 1. Attach to i the information currently found in the symbol table regarding the next use and liveness of x, y, and z. 2. In the symbol table, set x to "not live" and "no next use. 3. In the symbol table, set y and z to "live" and the next uses of y and z to i 36
37 Computing Liveness Information Start at the last statement in B and scan backwards At each statement i: x = y + z in B, we do the following: 1. Attach to i the information currently found in the symbol table regarding the next use and liveness of x, y, and z. 2. In the symbol table, set x to "not live" and "no next use. 3. In the symbol table, set y and z to "live" and the next uses of y and z to i x = 1 y = x + 3 z = x * 3 x = x * z can we change the order between 2 and 3? 37
38 common-subexpression elimination common-subexpression elimination a = b + c b = a d c = b + c d = a - d a = b + c b = a d c = b + c d = b 38
39 DAG Representation of Basic Blocks a = b + c b = a - d c = b + c d = a - d - + b,d c + a d0 b0 c0 39
40 DAG Representation of Basic Blocks a = b + c b = b - d c = c + d e = b + c + + e a - b + c b0 c0 d0 40
41 algebraic identities a = x^2 b = x*2 c = x/2 d = 1*x a = x*x b = x+x c = x*0.5 d = x 41
42 coming up next register allocation 42
43 The End 43
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