1、The Exercises of Chapter Six6.2应该在 numdigit 产生式中再加一条语义规则：numd.count=1 用来进行初始化。6.46.7 Consider the following grammar for simple Pascal-style declarations:delc var-list : typevar-list var-list, id | idtype integer | realWrite an attribute grammar for the type of a variable.Solution Grammar Rule Semant
2、ic Rulesdelc var-list : type var-list.type = type.typevar-list1 var-list2, id val-list2.type=var-list1.typeid.type=var-list1.typevar-list id id.type=var-list.typetype integer type.type= INTERGERtype real type.type=REAL6.10 a. Draw dependency graphs corresponding to each grammar rule of Example 6.14
3、(Page 283) , and for the expression 5/2/2.0.b. Describe the two passes required to compute the attributes on the syntax tree of 5/2/2.0, including a possible order in which the nodes could be visited and the attribute values computed at each point.c. Write pseudcode for procedures that would perform
4、 the computations described in part(b).Solutiona. The grammar rules of Example 6.14S expexp exp/exp | num | num.numThe dependency graphs for each grammar rule:S expval SisFloat etype val expexp exp / expisFloat etype val expisFloat etype val exp / isFloat etype val expexp numisFloat etype val expval
5、 numexp num.numisFloat etype val expval num.numThe dependency graphs for the expression: 5/2/2.0val SIsFloat etype val expisFloat etype val exp / isFloat etype val expisFloat etype val exp / isFloat etype val exp val num.num(2.0)val num val num(5) (2)b. The first pass is to compute the etype from is
6、Float.The second pass is to compute the val from etype.The possible order is as follows:val S122IsFloat etype3 val 11 exp1isFloat 4 etype val9 exp / isFloat etype val10 expisFloat 5etype val6 exp / isFloat 7etype val8 exp val num.num(2.0)val num val num(5) (2)c. The pseudcode procedure for the compu
7、tation of the isFloat.Function EvalisFloat(T: treenode): BooleanVar temp1, temp2: BooleanBeginCase nodekind of T ofexp: temp1= EvalisFloat(left child of T);if right child of T is not nil thentemp2=EvalisFloat( right child of T)return temp1 or temp2elsereturn temp1;num:return false;num.num:return tru
8、e;endFunction Evalval(T: treenode, etype:integer): VALUEVar temp1, temp2: VALUEBeginCase nodekind of T ofS:Return(Evalval(left child of T, etype);Exp: If etype=EMPTY thenIf EvalisFloat(T) then etype:=FLOAT;Else etype=INT;Temp1=Evalval(left child of T, etype)If right child of T is not nil thenTemp2=E
9、valval(right child of T, etype);If etype=FLOAT thenReturn temp1/temp2;ElseReturn temp1 div temp2;ElseReturn(temp1);Num:If etype=INTReturn(T.val);ElseReturn(T.val);Num.num:Return(T.val).6.11Dependency graphs corresponding to the numbered grammar rules in 6.4:Dependency graph for the string 3 *(4+5) *
10、6:6.21 Consider the following extension of the grammar of Figure 6.22(page 329) to include function declarations and calls:program var-decls;fun-decls;stmtsvar-decls var-decls;var-decl|var-declvar-decl id: type-exptype-exp int|bool|array num of type-expfun-decls fun id (var-decls):type-exp;bodybody
11、expstmts stmts;stmt| stmtstmt if exp then stmt | id:=expexp exp + exp| exp or exp | expexp|id(exps)|num|true|false|idexps exps,exp|expa. Devise a suitable tree structure for the new function type structure, and write a typeEqual function for two function types.b.Write semantic rules for the type che
12、cking of function declaration and function calls(represented by the rule exp id(exps),similar to rules of table 6.10(page 330).Solutiona. One suitable tree structure for the new function type structure:The typeEqual function for two function type:Function typeEqual-Fun(t1,t2 : TypeFun): BooleanVar t
13、emp : Boolean;p1,p2:TypeExpbeginp1:=t1.lchild;p2:=t2.lchild;temp:=true;while temp and p1nil dobegintemp=typeEqual-Exp(p1,p2);p1=p1.sibling;p2=p2.sibling;endif temp then return(typeEqual-Exp(t1.rchild,t2.rchild);return(temp);endb. The semantic rules for type checking of function declaration and funct
14、ion call:fun-decls fun id (var-decls):type-exp; bodyid.type.lchild:=var-decls.type;Fun(id)Type-exp.Type-expid.type.rchild:=type-exp.type;insert(id.name,id.typefun)exp id(exps)if isFunctionType(id.type) andtypeEqual-Exp(id.type.lchild,exps.type) thenexp.type=id.type.rchild;else type-error(exp)The exe
15、rcise of chapter seven7.2 Draw a possible organization for the runtime environment of the following C program, similar to that of Figure 7.4 (Page 354).a. After entry into block A in function f.b. After entry into block B in function g.int a10;char *s = “hello”Int f(int i, int b ) int j=i;A: int i=j
16、;Char c = bI;return 0;void g(char *s) char c=s0;B: int a5;main int x=1x = f(x,a);g(s);return 0;Solutiona. After entry into block A in function f.a9a8a7a6a5a4a3a2a1a0*(s+4): o *(s+3):l*(s+2):l*(s+1):e*s:hx: 1i: 1b9b8b7b6b5b4b3b2b1b0control linkreturn addressj:1i:1c:b1Global/static areaActivation reco
17、rd of mainActivation record of f after entering the Block Afpspb. After entry into block B in function g.a9a8a7a6a5a4a3a2a1a0*(s+4): o *(s+3):l*(s+2):l*(s+1):e*s:hx: 0*(s+4): o *(s+3):l*(s+2):l*(s+1):e*s:hcontrol linkreturn addressc: ha4a3a2a1a07.8 In languages that permit variable numbers of argume
18、nts in procedure calls, one way to find the first argument is to compute the arguments in reverse order, as described in section 7.3.1, page 361.a. One alternative to computing the arguments in reverse would be to reorganize the activation record to make the first argument available even in the pres
19、ence of variable arguments. Describe such an activation record organization and the calling sequence it would need.b. Another alternative to computing the arguments in reverse is to use a third point(besides the sp and fp), which is usually called the ap (argument pointer). Describe an activation re
20、cord structure that uses an ap to find the first argument and the calling sequence it would need.Solutiona. The reorganized activation record.Global/static areaActivation record of mainActivation record of g after entering the Block BfpspControl linkReturn addressArgument 1argument nLocal-varThe cal
21、ling sequence will be:(1) store the fp as the control link in the new activation record;(2) change the fp to point to the beginning of the new activation record;(3) store the return address in the new activation record;(4) compute the arguments and store their in the new activation record in order;(
22、5) perform a jump to the code of procedure to be called.b. The reorganized activation record.Argument 1argument n Control linkReturn addressLocal-varThe calling sequence will be:(1) set ap point to the position of the first argument.(2) compute the arguments and store their in the new activation rec
23、ord in order;(3) store the fp as the control link in the new activation record;(4) change the fp to point to the beginning of the new activation record;(5) store the return address in the new activation record;(6) perform a jump to the code of procedure to be called.7.15 Give the output of the follo
24、wing program(written in C syntax) using the four parameter methods discussed in section 7.5.#include int i=0;void p(int x, int y) x +=1;i +=1;y +=1;fpspfpspapmain int a2=1,1;p(ai, ai);printf(“%d %dn”, a0, a1);return 0;Solutionpass by value: 1, 1pass by reference: 3, 1pass by value-result: 2, 1pass by name: 2, 2
