Monday, November 7, Structures and dynamic memory

Transcription

1 Monday, November 7, 2016 Topics for today Structures Structures and dynamic memory Grammars and Languages (Chapter 7) String generation Parsing Regular languages Structures We have seen one composite data structure - the array. Many high level languages enable users to define structures. The main differences between arrays and structures are: elements of an array are accessed by numeric index, structure elements have names array elements are all the same type, structure elements can be of different types. In C Consider for example, the following C definition of a Person structure. typedef struct { char name[20]; int age; char dept[30]; int salary; } Person; With this definition we can declare and manipulate structures of type Person. Person A, B, C; input (A.age); B.salary = C.salary + 500; C.salary = C.age*4; We can also use Person to create larger data structures as in the following examples Example 1 Person Amgen[6000]; Amgen[k].age=30; Comp 162 Notes Page 1 of 11 November 7, 2016

2 Example 2 typedef struct { Person President; Person Faculty[80]; int enrollment; char URL[30]; } College; College CSUCI; Print(CSUCI.President.salary); Structures in Pep/8 (see p. 310) The easiest way to implement a structure in Pep/8 is to define constants representing the offsets within the structure where each field starts. Thus if we assume that a char is 1 byte and an int is 2 bytes, then in the example of Person name occupies bytes 0 through 19 age occupies bytes 20 and 21 dept occupies bytes 22 through 51 salary occupies bytes 52 and name age dept salary thus we will have name:.equate 0 age :.equate 20 dept:.equate 22 salary:.equate 52 The declarations of the Persons in our example are just blocks of the appropriate size A:.block 54 B:.block 54 C:.block 54 and the accessing/updating instructions in our example are as follows Comp 162 Notes Page 2 of 11 November 7, 2016

3 ; input (A.age) ; ldx age,i deci A,x ; B.salary = C.salary ; ldx salary,i lda C,x ; get C.salary adda 500,i sta B,x ; store in B.salary ; C.salary = C.age * 4 ; ldx age,i lda C,x ; C.age asla asla ; C.age * 4 ldx salary,i sta C,x ; to C.salary Structures and dynamic memory allocation We can use dynamic memory allocation, pointers and structures together to implement data structures such as linked lists and trees. In a high-level language, the structure representing a list node might consist of an integer data item and a pointer to the next node and be declared pictorially typedef struct { int data; node* next; } node; data next Warford's code (Fig. 6.47) shows how to construct a linked list. Each new item is prepended to the existing list. Thus if numbers were input in the order 4, 2, 9, 12, the resulting list could be depicted first Comp 162 Notes Page 3 of 11 November 7, 2016

4 The contents of the list can be output in the following manner. P = first while (P!= 0) // zero is our NULL pointer { output(p->data); P = P->next; } Here is a Pep/8 translation (shorter than Warford's code at the end of Fig. 6.47). We assume that both P and first are local variables stored on the stack with appropriate constants defined for symbols P and first. Think of each list node as an array with two elements (one data, one pointer) pointed to from the stack and you will see how mode sxf is the right one to use. lda first,s sta P,s test: breq done ; finished when pointer = 0 ldx data,i deco P,sxf ; output data field of structure pointed to by P charo ' ',i ; and a separating space ldx next,i lda P,sxf ; get field containing pointer to next node sta P,s ; and assign to P br test ; go see if finished done:... Comp 162 Notes Page 4 of 11 November 7, 2016

5 Grammars and Languages (Chapter 7) A (formal) language is a set of character strings. For example, the Java language is the set of all Java programs, the C language is the set of all C programs. Most interesting languages are, like Java and C, infinite sets of strings. An example of a finite language is the set of California license plates. We can represent a finite language by listing all the strings but we cannot do that for an infinite language and for large finite languages we also might like something more convenient. A grammar can be used to represent a language. A grammar can be used (a) to systematically generate the strings and (b) to determine if a given string is in the language the parsing problem For a particular language there might be more than one grammar that can represent it. A grammar has 4 components: (1) a set of terminal symbols (usually denoted with lower case) (2) a set of non-terminal symbols (denote using upper case) (3) a start symbol - one of the non-terminals, usually S (4) a set of replacement rules Example Grammar 1 (1) { x y z } (2) { A B C S } (3) S (4) S A BB A C B C x C y z A rule such as C y z is read as "C can be replaced by y or z" String Generation We start with the start symbol and apply rules until there are only terminal symbols in the string. The string is then one of the strings in the language. For example, one sequence of replacements for our example is Another is S BB CB Cx zx S A C z Comp 162 Notes Page 5 of 11 November 7, 2016

6 Our first example grammar represents a finite language. Here is the complete set of strings we can generate from this grammar { y z xx xy xz yx yy yz zx zy zz } Example Grammar 2 (omitting everything except the replacement rules) S ARTICLE ADJ NOUN ARTICLE the a ADJ red wild absent-minded NOUN desk puppy car The language defined by this grammar is also finite. Using this grammar we can generate 18 strings (sentences). All are syntactically correct (match the rules of the language) though not all make sense. [ Noam Chomsky gave colorless green ideas sleep furiously as an example of a sentence that is grammatically correct but meaningless] Example sentences are: the wild puppy the absent-minded desk a red car Infinite Languages We can generate an infinite language by using rules that are directly or indirectly recursive. An example of a recursive replacement rule is X ax a which causes X to be replaced by one or more a's, For example X ax aax aaax aaaa X a X ax aax aaax aaaax aaaaax aaaaaax aaaaaaa Here is another example of recursion ADJECTIVAL_PHRASE -> <empty> ADJECTIVE ADJECTIVAL_PHRASE ADJECTIVE -> blue fizzy tame Example strings generated from ADJECTIVAL_PHRASE are: tame blue blue fizzy fizzy tame tame blue Comp 162 Notes Page 6 of 11 November 7, 2016

7 The empty string (sometimes denoted Λ) is useful in defining languages. For example, if we want X to be zero or more a s X ax Λ The parsing problem The parsing problem is determining whether a particular string is in a particular language. It is usually of more interest than string generation. We are asking the question "is the string X in the language defined by grammar G"? A compiler needs to be able to tell if the input is a valid program as defined by the rules of the programming language grammar. We could start with the start symbol of G and see if we can generate the string X or we could start with X and see if we could collapse it back to the start symbol. Either way conceptually we try to build a "parse tree" with start symbol S at the root and the elements of string X as the leaves. Grammar String S WX W bw a b X XYZ Λ Y ay Λ Z bc bbaabc Parse tree (is there another?) S / \ W X bw X Y Z b ay bc ay bb aa bc Comp 162 Notes Page 7 of 11 November 7, 2016

8 Yes, there is another possible tree. S / \ W X bw X Y Z bw ay bc bba a bc So this particular grammar is ambiguous for at least one string there is more than one way to generate that string. Ambiguity is a property of the grammar not the language. There are also likely to be many grammars that represent a particular language. Regular languages Noam Chomsky defined a hierarchy of languages. The types of language are defined by what kind of replacement rules are permitted in the grammar. The fewer the restriction on the rules, the more powerful the device we need to solve the parsing problem. Regular languages (Type 3 in Chomsky s hierarchy) have the most restrictions on replacement rules. Rules in a regular grammar can only be of two forms: Non-terminal terminal Non-terminal terminal Non-terminal A regular language can be represented by a regular expression. A regular expression is defined as follows: any symbol is a regular expression if A and B are regular expressions then so are A* AB and A B A* means zero or more repetitions of A. AB is the concatenation of A and B, that is, A followed by B A B means "A or B" We can use parentheses also. In other words we can construct regular expressions using repetition, sequence and choice (note the similarity with program constructs). Comp 162 Notes Page 8 of 11 November 7, 2016

9 Example regular expressions letter (letter digit)* [ this is the set of identifiers ] (digit period digit* ) (digit* period digit) [ some real numbers ] identifier colon [ label in assembly language] open-paren closing-paren open-paren id-list closing-paren [ parameter list] identifier identifier "," id-list [ identifier list] An example of a language which is not regular is the set of well-formed parenthesis strings. This language includes (()) (()()()) ((()(()))) It is not possible to devise a regular expression that defines the set of well-formed parenthesis strings. This language is Type 2 in Chomsky s hierarchy. A grammar for it is S () S() ()S (S) Reading Finish Chapter 6. Which parts of Chapter 7 to read as we go through it? Section 7.1: You can skip the part on Context-Sensitive grammars pp and skim the part on C++ from pages 341 to the end of section 7.1. Section 7.2: you can skip the stuff on non-determinism through to the end of the section (pp ). Section 7.3: skip the material from An Input Buffer Class to the end of the section (pp ) Section 7.4. Skim this. We will have an alternative approach to code generation. Comp 162 Notes Page 9 of 11 November 7, 2016

10 Review Questions 1. Suppose we define a structure type car as typedef struct = { int yearmade; char make[10]; int miles; int hp; char VIN[17]; } car; and we have an array of these as in car Carmax[500]; (a) How many bytes does Carmax occupy? (b) If, in the Pep/8 implementation, M is the address of a particular car, what is the address of the next one? (c) Write code that output the number of old cars (made before 2005). 2. Turn our list-printing loop into a recursive subroutine that takes a pointer to a list as its parameter. 3. Consider the following grammar S aas Saaa Λ Identify a string that can be generated in more than one way 4. Give a grammar that represents the same language as the regular expression a*bcc*. 5. Give a regular expression that represents the same language as the following grammar. S P Q P kp k Q Λ QR R a b Comp 162 Notes Page 10 of 11 November 7, 2016

11 Review Answers 1. (a) 500 * ( ) = 500 * 33 = (b) M+33 (c) ldx 0,i loop:lda carmax,x cpa 2005,i brge skip lda total,d adda 1,i sta total,d skip:addx 33,i cpx 16500,i brlt loop deco total 2. Here is a possible answer Plist: lda 2,s ; parameter breq exit ; branch if null list ldx data,i deco 2,sxf; output P.data subsp 2,i ; for parameter of recusrion ldx next,i lda 2,sxf ; P.next sta 0,s ; is parameter call Plist; of recursion addsp 2,I ; done with parameter exit: ret0 3. For example S aas aaaas aaaaaas aaaaaa S Saaa Saaaaaa aaaaaa 4. For example 5. kk* (a b)* S PQR P Λ ap Q b R c cr Comp 162 Notes Page 11 of 11 November 7, 2016