编译原理与实践 第六 七章 答案

The Exercises of Chapter Six6.2应该在nu m→digit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 Semantic 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 (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 the computations described in part(b). [Solution]a. 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 / expvalexp →numisFloat etype val expval numexp →num.numisFloat etype val expval num.numThe dependency graphs for the expression: 5/2/2.0val SexpisFloat val num.num(2.0)val num val num(5)(2)b. The first pass is to compute the etype from isFloat.The second pass is to compute the val from etype.The possible order is as follows:val S1221i expisFloat val num.num(2.0)(5)(2)c. The pseudcode procedure for the computation 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 true;endFunction Evalval(T: treenode, etype:integer): V ALUEVar temp1, temp2: V ALUEBeginCase 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=Evalval(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) *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 → expstmts → stmts;stmt| stmtstmt → if exp then stmt | id:=expexp → exp + exp| exp or exp | exp[exp]|id(exps)|num|true|false|idexps→ exps,exp|expa.Devise a suitable tree structure for the new function type structure, and write a typeEqualfunction for two function types.b.Write semantic rules for the type checking of function declaration and functioncalls(represented by the rule exp →id(exps)),similar to rules of table 6.10(page 330).[Solution]a. One suitable tree structure for the new function type structure:The typeEqual function for two function type:Function typeEqual-Fun(t1,t2 : TypeFun): BooleanVar temp : Boolean;p1,p2:TypeExpbeginp1:=t1.lchild;p2:=t2.lchild;temp:=true;while temp and p1<>nil and p2<>nil 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 function call:fun-decls → fun id (var-decls):type-exp; bodyid.type.lchild:=var-decls.type;id.type.rchild:=type-exp.type;insert(,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 exercise 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.[Solution]a. Global/static area Activation record of main Activation record of f after entering the Block Afpsp7.8 In languages that permit variable numbers of arguments 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 presence 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 record structure that uses an ap to find the first argument and the calling sequence it would need.[Solution]a. The reorganized activation record.Global/static areaActivation record of mainActivation record of g after entering the Block BThe calling 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;(5)perform a jump to the code of procedure to be called.The 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 record 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 following program(written in C syntax) using the four parameter methods discussed in section 7.5.#include <stdio.h>int i=0;void p(int x, int y){ x +=1;i +=1;y +=1;}main{ int a[2]={1,1};p(a[i], a[i]);printf(“%d %d\n”, a[0], a[1]);return 0;}[Solution]pass by value: 1, 1pass by reference: 3, 1pass by value-result: 2, 1pass by name: 2, 2。