1. Cse322, Programming Languages and Compilers 1 6/22/2015 Lecture #4, April 12, 2007 Strings (representation, byte operation, copying), Structures (representation, anonymous value, field information, layout), Control Flow (basic blocks, generating code, loops).

2. Cse322, Programming Languages and Compilers 2 6/22/2015 Assignments Reading –Read chapter 7 sections 7.9 7.10 and 7.11 –Possible Quiz Monday on the reading.

3. Cse322, Programming Languages and Compilers 3 6/22/2015 Strings Strings are usually represented as byte sequences Operations on strings do not generally map onto hardware operations. –Load instructions load whole words –Strings are composed of bytes –Shifting and masking are often necessary String representations are often both language and machine dependent. –In C strings are null terminated adjacent arrays of char –In Java strings are byte arrays with their length stored explicitly.

4. Cse322, Programming Languages and Compilers 4 6/22/2015 Representation abc\0abc3 abc 3 Null terminated Length Prefixed Length plus pointer 2 Note that sharing is possible with the length plus pointer

5. Cse322, Programming Languages and Compilers 5 6/22/2015 Assignment of individual characters A[ 1 ] = b[ 2 ] loadI @b => r b cloadAI r b,2 => r 2 loadI @a => r a cstoreAI r 2 => r a,1 This is only possible if the machine has byte level load and store. Many machines do not.

6. Cse322, Programming Languages and Compilers 6 6/22/2015 Without byte oriented operations Masking and shifting are necessary without byte oriented operations. Masking –A mask is a word where “1”s are in the important positions and “0”s are in other positions. –For example in a 32 bit word, the mask for the second byte »00000000 00000000 11111111 00000000 » Ox0000FF00 in hex »Anding a mask with a word “zeros” out the unmasked bits andI 00000000 00000000 11111111 00000000 01011101 11011101 01010001 11110101 -> 00000000 00000000 01010001 00000000 Shifting –Shifting moves the bits over Shift 00000000 00000000 11111111 00000000,8 -> 00000000 00000000 00000000 11111111

7. Cse322, Programming Languages and Compilers 7 6/22/2015 a[ 1 ] = b[ 2 ] Load source word ( b ) –01011101 11011101 01010001 11110101 Mask away unwanted characters (every thing but 2) –00000000 00000000 01010001 00000000 Shift to byte position in word of target (position 1) – 00000000 00000000 00000000 01010001 Load target word (a) –01110100 10111011 00001011 11010111 Mask away the position of the target character –01110100 10111011 00001011 00000000 Or with shifted & masked source with masked target Or 00000000 00000000 00000000 01010001 01110100 10111011 00001011 00000000 -> 01110100 10111011 00001011 01010001 Store result in target address

8. Cse322, Programming Languages and Compilers 8 6/22/2015 Longer words If a and b are longer strings (longer than 4 characters) then we need to select the right word from the longer string. A[n] = B[m] The correct source word is ( n `div` 4 ) The correct source position in that word is ( n `mod` 4 ) Similar for target string A

9. Cse322, Programming Languages and Compilers 9 6/22/2015 Copying Strings. To copy a string we need to copy all the component characters. With byte oriented load and store this is easy With word oriented load and store again need to load and move words. –How many words must we move? –When do we need to mask? »How is this affected by length? »Word alignment of the two strings? Error conditions. –Since strings are generally allocated once. –A := B could cause an error if B is longer than A –Test for lengths, first.

10. Cse322, Programming Languages and Compilers 10 6/22/2015 String Concatenation A^B Compute lengths of A and B len A and len B Allocate (len A + len B ) bytes plus room for length and any alignment necessary. Copy A to target Copy B to target Set the length convention appropriately.

11. Cse322, Programming Languages and Compilers 11 6/22/2015 String Length Here we use the explicit information stored with the string. Null terminated –Loop and count until 0 is encountered Length Prefixed –Address of string stores the length Length plus pointer –Address of string stores length abc\0abc3 abc 3

12. Cse322, Programming Languages and Compilers 12 6/22/2015 Structures Structures are heterogeneous aggregates with statically known accessors. Statically known means we know their “offset” at compile time. Sometimes these are named. X.age Sometimes the names are implicit as in pattern matching in ML fun f (Node(x,y,z)) = … Positions of x, y, and z, are statically known Examples include C - struct struct node { Int value; Struct node *next; } Java - Objects with instance variables ML - datatypes with constructors with more than one field

13. Cse322, Programming Languages and Compilers 13 6/22/2015 Problems Anonymous values struct node { Int value; Struct node *next; } Node x f( *(X.next) ) Note that (X.next) is an anonymous value. A value without a name. Structure Layout –Layout requires alignment –Computing offsets for each field. –Offset depends on size of preceeding fields in the structure

14. Cse322, Programming Languages and Compilers 14 6/22/2015 Anonymous values Aliasing is a problem with anonymous values. –Pointers int a, *b; b = &a; –Array References Are x[i] and x[j-n] different? p1 = (node *) malloc(sizeof(node)); p2 = (node *) malloc(sizeof(node)); If (...) then p3 = p1; else p3 = p2; p1->value =... p2->value =... w := p1->value; It is clear that p1->value is stored in a register. But what register? It depends upon the path through the if then else. Anonymous values are often stored in memory because we can’t tell when they might change because of aliasing

15. Cse322, Programming Languages and Compilers 15 6/22/2015 Recording and using field information Structure name Field namelengthoffsettype node2 fields value4 bytes0int next4 bytes4node * struct node { int value; struct node *next; } p1->next loadI 4 => r 1 // offset of next loadAI r p1,r 1 => r 2 // value of p1->next

16. Cse322, Programming Languages and Compilers 16 6/22/2015 Layout of structures When laying out structures –Meet all alignment rules –Minimize the amount of space used –Statically know the offset of each field. Struct example { int fee; double fie; int foe; double fum; } e1; Structure name Field name lengthNaïve offset type example4 fields fee4 bytes0int fie8 bytes4double foe4 bytes12int fum8 bytes16double fee…fiefoe …fum 0 4 8 16 24 Note that the alignment of fie on double word boundaries makes naïve offset be incorrect

17. Cse322, Programming Languages and Compilers 17 6/22/2015 Alternate structure We can reorder the layout of the fields As long as the table is correct, the programmer cannot observe this change. This also save space as we don’t use unnecessary padding Structure name Field name lengthoffsettype example4 fields fee4 bytes16int fie8 bytes0double foe4 bytes20int fum8 bytes8double fee fiefoe fum 0 8 16 20

18. Cse322, Programming Languages and Compilers 18 6/22/2015 Arrays of structures struct node { int value; int age; } node x[4]; Value = 5 Age = 34 Value = 2 Age = 18 Value = 0 Age = 3 Value = 9 Age = 45 01230123 5 2 0 9 34 18 3 45 Value Age We can represent these in at least two ways. Performance may vary.

19. Cse322, Programming Languages and Compilers 19 6/22/2015 Unions and run-time tags Unions can have several different layouts at runtime. In order to distinguish at runtime, the user must add a tag field that can be tested at runtime to distinguish. struct two { int tag; union choice { struct { char * name } A struct { int age } B } field } u2; In ML the tags are the constructor names!

20. Cse322, Programming Languages and Compilers 20 6/22/2015 Basic Blocks A (maximal length) straight- line code segment. Any jump or label (because it is the target of a jump) ends a basic block. loadI @a => r2 loadAO rA,r2 => r3 loadI @b => r4 loadAO rA,r4 => r5 L1: comp r3,r5 => cc1 cbr_Lt cc1 -> L2,L5 L5: loadI @c => r6 loadAO rA,r6 => r7 loadI @d => r8 loadAO rA,r8 => r9 comp r7,r9 => cc2 cbr_Lt cc2 -> L6,L3 L6: loadI @e => r10 loadAO rA,r10 => r11 loadI @f => r12 loadAO rA,r12 => r13 comp r11,r13 => cc3 cbr_Lt cc3 -> L2,L3 L2: loadI true => r1 jumpI -> L4 L3: loadI false => r1 jumpI -> L4 L4: nop

21. Cse322, Programming Languages and Compilers 21 6/22/2015 Sources Basic blocks are produced by –Control Flow constructs in the language »If-then-else »Loops –Positional evaluation of booleans –Short circuit evaluation

22. Cse322, Programming Languages and Compilers 22 6/22/2015 Predication vs jumps Recall the predicated move Mov_GT cc,r1,r2, => r3 Mostly we use these to avoid branching or jumps –if x<y then a <- c+d else a <- e+f –comp rx,ry => cc1 –add rc,rd => r1 –add re,rf => r2 –mov_LT cc1,r1,r2 => ra If the branches to the else and then are large, we may do too much speculative execution, so using jumps may be better. Other considerations –Expected frequency of one path over another –Complicated control flow (other if-then-else) inside the then or else

23. Cse322, Programming Languages and Compilers 23 6/22/2015 Generating code Because of our experience with short-circuit evaluation we have all the tools to generate code with control flow. We will need one more IR instruction datatype IR = LoadI of (string * Reg) | LoadAO of (Reg * Reg * Reg) | Arith of (Op * Reg * Reg * Reg) | Comp of (Reg * Reg * CC) | Neg of (Reg * Reg) | Cmp of (Op * Reg * Reg * Reg) | Cbr of (Op * CC * Label * Label) | JumpI of Label | Lab of (Label * IR) | Nop | StoreAO of (Reg * Reg * Reg)

24. Cse322, Programming Languages and Compilers 24 6/22/2015 Translating statements fun stmt dict x = case x of Assign (NONE,v,NONE,exp) => let val result = expr dict exp val b = base (Var(NONE,v)) val delta = offset (Var(NONE,v)) in emit (StoreAO(result,b,delta)); result end

25. Cse322, Programming Languages and Compilers 25 6/22/2015 z := x – (2 * y) loadI @x => r1 loadAO rA,r1 => r2 loadI 2 => r3 loadI @y => r4 loadAO rA,r4 => r5 Mul r3,r5 => r6 Sub r2,r6 => r7 loadI @z => r8 storeAO r7 => rA,r8

26. Cse322, Programming Languages and Compilers 26 6/22/2015 If then else fun stmt dict x = case x of If (tst,thenS,elseS) => let val [start,thenL,elseL,endL] = NextLabel 4 in short dict tst start thenL elseL; emitAt thenL Nop; stmt dict thenS; emit (JumpI,endL); emitAt elseL Nop; stmt dict elseS; emitAt endL Nop; end; Note how we take advantage of the short circuit evaluation mechanism

27. Cse322, Programming Languages and Compilers 27 6/22/2015 Loops Loops have multiple parts –Initialization –Tests for termination –Body –Jump to continue loop Your homework on tuesday will be to extend S04code.sml to include translation of the while statement.