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Grammer: lexp→atom | list atom→number | identifier list→( lexp-seq) lexp-seq→lexp-seq lexp | lexp

a. Write C type declarations for the abstract syntax tree structure for the grammar.
b. Translate the grammar into a EBNF
c. Draw the syntax diagram for the EBNF

User Brahadeesh
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1 Answer

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Hi! I'm happy to help you with your question.

a. C type declarations for the abstract syntax tree structure for the grammar:

```c
typedef enum { ATOM_NUMBER, ATOM_IDENTIFIER } AtomType;

typedef struct Atom {
AtomType type;
union {
int number;
char *identifier;
} value;
} Atom;

typedef struct LExp {
enum { ATOM_NODE, LIST_NODE } type;
union {
Atom atom;
struct LExpSeq *list;
} value;
} LExp;

typedef struct LExpSeq {
LExp *lexp;
struct LExpSeq *next;
} LExpSeq;
```

b. EBNF translation of the grammar:

```
lexp = atom | list ;
atom = number | identifier ;
list = "(", lexp-seq, ")" ;
lexp-seq = {lexp}, lexp ;
```

c. Unfortunately, I'm unable to draw syntax diagrams as a text-based AI. However, I can describe the process for you, and you can use an online tool or draw it manually.

1. Start with a rectangle for each non-terminal (lexp, atom, list, lexp-seq).
2. Connect each non-terminal to its respective production rules using arrows.
3. Use rounded rectangles for terminals (number, identifier, "(", ")").
4. For the "|" operator, draw two branches, one for each alternative.
5. For the "{}" operator, draw a loop from the start to the end of the repeated sequence.

I hope this helps! Let me know if you have any further questions.

User Kacalapy
by
7.7k points