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Theory of Compilation 236360
Erez Petrank
Lecture 10: Code Generation
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Executable
code
exe
Source
text
txt
Compiler
Lexica
lA
naly
sis
Syn
tax
An
aly
sis
Sem
an
ticA
naly
sis
Inte
r.R
ep
. (IR)
Cod
e
Gen
.
chara
cters
toke
ns A
ST
(Abstra
ct Synta
x Tre
e)
An
nota
ted
AS
T
3
target languages
Absolute machine code
Code
Gen.
Relativemachine code
Assembly
IR + Symbol Table
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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
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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)
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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 – data/general 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)
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Not all registers are born equal
EAX Required operand of MUL,IMUL,DIV and IDIV instructions Contains the result of these operations
EDX 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
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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
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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)
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Base Displacement Addressing
Mov (%ecx,%ebx,4), %eax
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Array Base Reference
4 4
0 2 4 5 6 7 1
4 4 4 4 4 4
%ecx = base%ebx = 3
offset = base + (index*scale) + displacement
offset = base + (3*4) + 0 = base + 12
(%ecx,%ebx,4)
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Basic Blocks
An important notion. Start by breaking the IR into basic blocks A 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
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Example
False
B1
B2 B3
B4
True
t1 := 4 * it2 := a [ t1 ]if t2 <= 20 goto B3
t5 := t2 * t4
t6 := prod + t5
prod := t6
goto B4
t7 := i + 1i := t2
Goto B5
t3 := 4 * it4 := b [ t3 ]goto B4
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creating basic blocks
Input: A sequence of three-address statements Output: A list of basic blocks with each three-
address 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
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control flow graph
A directed graph G=(V,E) nodes V = basic blocks edges E = control flow
(B1,B2) E if control from B1 flows to B2
A loop is a strongly connected component of the graph that has a single entry point.
An inner loop is a loop that has no sub-loop.
B1
B2
t1 := 4 * it2 := a [ t1 ]t3 := 4 * it4 := b [ t3 ]t5 := t2 * t4
t6 := prod + t5
prod := t6
t7 := i + 1i := t7
if i <= 20 goto B2
prod := 0i := 1
example
1) i = 1
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)
12) i=1
13) t5=i-1
14) t6=88*t5
15) a[t6]=1.0
16) i=i+1
17) if I <=10 goto (13)
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i = 1
j = 1
t1 = 10*It2 = t1 + jt3 = 8*t2t4 = t3-88a[t4] = 0.0j = j + 1if j <= 10 goto B3i=i+1if i <= 10 goto B2
i = 1
t5=i-1t6=88*t5a[t6]=1.0i=i+1if I <=10 goto B6
B1
B2
B3
B4
B5
B6
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;
source IR
CFG
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Optimizations Possible to obtain performance improvements by
working inside basic block boundaries. A better accuracy is obtained when considering
all blocks in a routine. It is best to consider all blocks in the program
(whole program analysis). But: Costly Usually unknown.
Any optimization must start with program “understanding” = program analysis.
An example: to allocate registers efficiently we must analyzing liveness of variables.
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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
If x is defined in instruction i, used in instr. j, and there is a computation path from i to j that does not modify x, then instr. j is using the value of x that is defined in instr. i.
y = 42z = 73
x = y + zprint(x);
x is live, y dead, z dead
x undef, y live, z live
x undef, y live, z undef
x is dead, y dead, z dead
(showing state after the statement)
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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
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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.
Algorithm: 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
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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
can we change the order between 2 and 3?
x = 1 y = x + 3 z = x * 3 x = x * z
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common-subexpression elimination
common-subexpression elimination
Easily identified by DAG representation.
a = b + cb = a - dc = b + cd = a - d
a = b + cb = a - dc = b + cd = b
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DAG Representation of Basic Blocks
a = b + cb = a - d
c = b + cd = a - d
b0 c0
+ d0
-
+
a
b,d
c
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DAG Representation of Basic Blocks
a = b + cb = b - dc = c + de = b + c
b0 c0
+
d0
- +a b c
+ e
• Also: discover dead code.
• Perform dead code elimination.
25
algebraic identities
a = x^2b = x*2c = x/2d = 1*x
a = x*xb = x+xc = x*0.5d = x
simple code generation
registers used as operands of instructions can be used to store temporary results can (should) be used as loop indexes due to frequent
arithmetic operation used to manage administrative info (e.g., runtime
stack)
number of registers is limited need to allocate them in a clever way
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simple code generation
assume machine instructions of the form LD reg, mem ST mem, reg OP reg,reg,reg
further assume that we have all registers available for our use ignore registers allocated for stack management
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simple code generation
translate each 3AC instruction separately
A register descriptor keeps track of the variable names whose current value is in that register. we use only those registers that are available for local use within a
basic block, we assume that initially, all register descriptors are empty.
As code generation progresses, each register will hold the value of zero or more names.
For each program variable, an address descriptor keeps track of the location or locations where the current value of that variable can be found. The location may be a register, a memory address, a stack
location, or some set of more than one of these Information can be stored in the symbol-table entry for that
variable28
simple code generation
For each three-address statement x := y op z,
1. Invoke getreg (x := y op z) to select registers Rx, Ry, and Rz.
2. If Ry does not contain y, issue: “LD Ry, y’ ”, for a location y’ of y.
3. If Rz does not contain z, issue: “LD Rz, z’ ”, for a location z’ of z.
4. Issue the instruction “OP Rx,Ry,Rz”
5. Update the address descriptors of x, y, z, if necessary. Rx is the only location of x now, and
Rx contains only x (remove Rx from other address descriptors).
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updating descriptors
1. For the instruction LD R, xa) Change the register descriptor for register R so it holds only x.b) Change the address descriptor for x by adding register R as an additional
location.
2. For the instruction ST x, R change the address descriptor for x to include its own memory location.
3. For an operation such as ADD Rx, Ry, Rz, implementing a 3AC instruction x = y + za) Change the register descriptor for Rx so that it holds only x.b) Change the address descriptor for x so that its only location is Rx. Note
that the memory location for x is not now in the address descriptor for x.c) Remove Rx from the address descriptor of any variable other than x.
4. When we process a copy statement x = y, after generating the load for y into register Ry, if needed, and after managing descriptors as for all load statements (rule 1):a) Add x to the register descriptor for Ry.b) Change the address descriptor for x so that its only location is Ry .
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example
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t= A – Bu = A- Cv = t + u
A = DD = v + u
A B C D = live outside the blockt,u,v = temporaries in local storate
R1 R2 R3
A B C
A B C
D
D t u v
t = A – B LD R1,A LD R2,B SUB R2,R1,R2
A t
R1 R2 R3A,R1 B C
A B C
DR2
D t u v
u = A – C LD R3,C SUB R1,R1,R3
v = t + u ADD R3,R2,R1
u t C
R1 R2 R3
A B C,R3
A B C
DR2
R1
D t u v
u t v
R1 R2 R3
A B C
A B C
DR2
R1
D t uR3
v
example
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t= A – Bu = A- Cv = t + u
A = DD = v + u
A B C D = live outside the blockt,u,v = temporaries in local storate
A = D LD R2, D
u A,D v
R1 R2 R3
R2 B C
A B CD,R2
R1
D t uR3
v
D = v + u ADD R1,R3,R1
exit ST A, R2 ST D, R1
D A v
R1 R2 R3R2
B C
A B CR1
D t uR3
v
u t v
R1 R2 R3
A B C
A B C
DR2
R1
D t uR3
v
D A v
R1 R2 R3A,R2 B C
A B CD,R1
D t uR3
v
33
design of getReg
many design choices
simple rules If y is currently in a register, pick a register
already containing y as Ry. No need to load this register
If y is not in a register, but there is a register that is currently empty, pick one such register as Ry
complicated case y is not in a register, but there is no free
register
34
design of getReg
instruction: x = y + z y is not in a register, no free register let R be a taken register holding value of a
variable v possibilities:
if the value v is available somewhere other than R, we can allocate R to be Ry
if v is x, the value computed by the instruction, we can use it as Ry (it is going to be overwritten anyway)
if v is not used later, we can use R as Ry otherwise: spill the value to memory by ST v,R
35
global register allocation
so far we assumed that register values are written back to memory at the end of every basic block
want to save load/stores by keeping frequently accessed values in registers e.g., loop counters
idea: compute “weight” for each variable for each use of v in B prior to any definition of v add 1 point for each occurrence of v in a following block using v add 2
points, as we save the store/load between blocks cost(v) = Buse(v,B) + 2*live(v,B)
use(v,B) is is the number of times v is used in B prior to any definition of v
live(v, B) is 1 if v is live on exit from B and is assigned a value in B after computing weights, allocate registers to the “heaviest”
values
36
Examplea = b + cd = d - be = a + f
bcdf
f = a - d
acde
cdef
b = d + fe = a – c
acdf
bcdef
b = d + c
cdef
bcdef
b,c,d,e,f live
B1
B2 B3
B4
acdef
cost(a) = B use(a,B) + 2*live(a,B) = 4cost(b) = 6cost(c) = 3cost(d) = 6cost(e) = 4cost(f) = 4
b,d,e,f live
37
ExampleLD R3,c
ADD R0,R1,R3SUB R2,R2,R1
LD R3,fADD R3,R0,R3
ST e, R3
SUB R3,R0,R2ST f,R3
LD R3,fADD
R1,R2,R3LD R3,c
SUB R3,R0,R3ST e, R3
LD R3,cADD R1,R2,R3
B1
B2 B3
B4
LD R1,bLD R2,d
ST b,R1ST d,R2
ST b,R1ST a,R2
38
Register Allocation by Graph Coloring
Address register allocation by liveness analysis reduction to graph coloring optimizations by program transformation
Main idea register allocation = coloring of an interference
graph every node is a variable edge between variables that “interfere” = are both
live at the same time number of colors = number of registers
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Example
v1
v2
v3
v4
v5
v6
v7
v8
time
V1
V8
V2
V4
V7
V6
V5
V3
40
Example
a = read();b = read();c = read();a = a + b + c;if (a<10) { d = c + 8; print(c);} else if (a<2o) { e = 10; d = e + a; print(e);} else { f = 12; d = f + a; print(f);} print(d);
a = read();b = read();c = read();a = a + b + c;if (a<10) goto B2 else goto B3
d = c + 8;print(c);
if (a<20) goto B4 else goto B5
e = 10;d = e + a;print(e);
f = 12;d = f + a;print(f);
print(d);
B1
B2B3
B4B5
B6
b
ac
d
ef
dd
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Example: Interference Graph
f a b
d e c
a = read();b = read();c = read();a = a + b + c;if (a<10) goto B2 else goto B3
d = c + 8;print(c);
if (a<20) goto B4 else goto B5
e = 10;d = e + a;print(e);
f = 12;d = f + a;print(f);
print(d);
B2B3
B4B5
B6
b
ac
d
ef
dd
42
Register Allocation by Graph Coloring
variables that interfere with each other cannot be allocated the same register
graph coloring classic problem: how to color the nodes of a graph
with the lowest possible number of colors bad news: problem is NP-complete (to even
approximate) good news: there are pretty good heuristic
approaches
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Heuristic Graph Coloring
idea: color nodes one by one, coloring the “easiest” node last
“easiest nodes” are ones that have lowest degree fewer conflicts
algorithm at high-level find the least connected node remove least connected node from the graph color the reduced graph recursively re-attach the least connected node
44
Heuristic Graph Coloring
f a b
d e c
f a
d e c
f a
d e
f
d e
stack: stack: b
stack: cb stack: acb
45
f
d e
stack: acb
f
d
stack: eacb
f
stack: deacbstack: fdeacb
f1
stack: deacb
f1
d2
stack: eacb
f1
d2 e1
stack: acb
f1 a2
d2 e1
stack: cb
Heuristic Graph Coloring
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f1 a2 b3
d2 e1 c1
f1 a2
d2 e1 c1
f1 a2
d2 e1
stack:
stack: bstack: cb
Heuristic Graph Coloring
Result:3 registers for 6 variables
Can we do with 2 registers?
47
two sources of non-determinism in the algorithm choosing which of the (possibly many) nodes of
lowest degree should be detached choosing a free color from the available colors
Heuristic Graph Coloring
48
The above heuristic gives a coloring of the graph. But what we really need is to color the graph with a
given number of colors = number of available registers. Many times this is not possible.
(Telling whether it is possible is NP-Hard.)
We’d like to find the maximum sub-graph that can be colored.
Vertices that cannot be colored will represent variables that will not be assigned a register.
Heuristic Graph Coloring
Similar Heuristic
1. Iteratively remove any vertex whose degree < k (with all of its edges).
2. Note: no matter how we color the other vertices, this one can be colored legitimately!
V1
V8
V2
V4
V7
V6
V5
V3
.4Now all vertices are of degree >=k (or graph is empty).5If graph empty: color the vertices one-by-one as in previous
slides. Otherwise ,.6Choose any vertex, remove it from the graph. Implication:
this variable will not be assigned a register. Repeat this step until we have a vertex with degree <k and go back to (1) .
Similar Heuristic
1. Iteratively remove any vertex whose degree < k (with all of its edges).
2. Note: no matter how we color the other vertices, this one can be colored legitimately!
V1
V8
V2
V4
V7
V6
V5
V3
.4Now all vertices are of degree >=k (or graph is empty).5If graph empty: color the vertices one-by-one as in previous
slides. Otherwise ,.6Choose any vertex, remove it from the graph. Implication:
this variable will not be assigned a register. Repeat this step until we have a vertex with degree <k and go back to (1) .
Source of non-determinism: choose which vertex to remove in (6).This decision determines the number of spills.
51
Summary: Code Generation
Depends on the target language and platform. GNU Assembly IA-32 platform.
Basic blocks and control flow graph show program executions paths.
Determining variable liveness in a basic block. useful for many optimizations. Most important use: register allocation.
Simple code generation. Better register allocation via graph coloring
heuristic.
