编译原理与实践 第六 七章 答案.doc

The Exercises of Chapter Six 6.2 应该在num→digit产生式中再加一条语义规则：numd.count=1用来进行初始化。 6.4 6.7 Consider the following grammar for simple Pascal-style declarations: delc → var-list : type var-list → var-list, id | id type → integer | real Write an attribute grammar for the type of a variable. [Solution] Grammar Rule Semantic Rules delc → var-list : type var-list.type = type.type var-list1 → var-list2, id val-list2.type=var-list1.type id.type=var-list1.type var-list → id id.type=var-list.type type → integer type.type= INTERGER type → real type.type=REAL 6.10 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 6.14 S → exp exp → exp/exp | num | num.num The dependency graphs for each grammar rule: S → exp val S isFloat etype val exp exp → exp / exp isFloat etype val exp isFloat etype val exp / isFloat etype val exp exp → num isFloat etype val exp val num exp → num.num isFloat etype val exp val num.num The dependency graphs for the expression: 5/2/2.0 val S IsFloat etype val exp isFloat etype val exp / isFloat etype val exp isFloat etype val exp / isFloat etype val exp val num.num (2.0) val num val num (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 S12 2IsFloat etype3 val 11 exp 1 isFloat 4 etype val9 exp / isFloat etype val10 exp isF