任务
给定一条特定的Sum语言语句, 将它编译成Stack语言的语句.
在Sum语言中,语句由数字或数字与+组成,例如
- 1
- 1 + 2
- 1 + 2 + 3
由于加法满足左结合,因此第三条式子可理解为 ((1 + 2) + 3).
而在Stack语言中存在两个操作行为,分别是Push, Add, 具体为
Push x; x表示整型数字。
Add : 从栈顶弹出两个数字,相加后再次把结果压入栈顶,可把*Add用以下四个步骤来描述:
- Pop a
- Pop b
- c = a + b
- Push c
程序
#include <stdio.h>
#include <stdlib.h>
#define TODO() \
do{ \
printf ("\nAdd your code here: file \"%s\", line %d\n", \
__FILE__, __LINE__); \
}while(0)
///////////////////////////////////////////////
// Data structures for the Sum language.
enum Exp_Kind_t {EXP_INT, EXP_SUM};
struct Exp_t
{
enum Exp_Kind_t kind;
};
struct Exp_Int
{
enum Exp_Kind_t kind;
int i;
};
struct Exp_Sum
{
enum Exp_Kind_t kind;
struct Exp_t *left;
struct Exp_t *right;
};
// "constructors"
struct Exp_t *Exp_Int_new (int i)
{
struct Exp_Int *p = malloc (sizeof(*p));
p->kind = EXP_INT;
p->i = i;
return (struct Exp_t *)p;
}
struct Exp_t *Exp_Sum_new (struct Exp_t *left, struct Exp_t *right)
{
struct Exp_Sum *p = malloc (sizeof(*p));
p->kind = EXP_SUM;
p->left = left;
p->right = right;
return (struct Exp_t *)p;
}
// "printer"
void Exp_print (struct Exp_t *exp)
{
switch (exp->kind){
case EXP_INT:{
struct Exp_Int *p = (struct Exp_Int *)exp;
printf ("%d", p->i);
break;
}
case EXP_SUM:{
struct Exp_Sum *p = (struct Exp_Sum *)exp;
Exp_print (p->left);
printf ("+");
Exp_print (p->right);
break;
}
default:
break;
}
}
//////////////////////////////////////////////
// Data structures for the Stack language.
enum Stack_Kind_t {STACK_ADD, STACK_PUSH};
struct Stack_t
{
enum Stack_Kind_t kind;
};
struct Stack_Add
{
enum Stack_Kind_t kind;
};
struct Stack_Push
{
enum Stack_Kind_t kind;
int i;
};
// "constructors"
struct Stack_t *Stack_Add_new ()
{
struct Stack_Add *p = malloc (sizeof(*p));
p->kind = STACK_ADD;
return (struct Stack_t *)p;
}
struct Stack_t *Stack_Push_new (int i)
{
struct Stack_Push *p = malloc (sizeof(*p));
p->kind = STACK_PUSH;
p->i = i;
return (struct Stack_t *)p;
}
/// instruction list
struct List_t
{
struct Stack_t *instr;
struct List_t *next;
};
struct List_t *List_new (struct Stack_t *instr, struct List_t *next)
{
struct List_t *p = malloc (sizeof (*p));
p->instr = instr;
p->next = next;
return p;
}
// "printer"
void List_reverse_print (struct List_t *list)
{
TODO();
struct List_t *p = list;
while (p)
{
switch (p->instr->kind)
{
case STACK_ADD:{
printf("Add\n");
break;
}
case STACK_PUSH:{
struct Stack_Push *comm = (struct Stack_Push *)p->instr;
printf("Push %d\n", comm -> i);
break;
}
default:
break;
}
p = p -> next;
}
}
//////////////////////////////////////////////////
// a compiler from Sum to Stack
struct List_t *all = 0;
void emit (struct Stack_t *instr)
{
all = List_new (instr, all);
}
void compile (struct Exp_t *exp)
{
switch (exp->kind){
case EXP_INT:{
struct Exp_Int *p = (struct Exp_Int *)exp;
emit (Stack_Push_new (p->i));
break;
}
case EXP_SUM:{
TODO();
struct Exp_Sum *q = (struct Exp_Sum *)exp;
compile(q->left);
compile(q->right);
emit(Stack_Add_new());
break;
}
default:
break;
}
}
//////////////////////////////////////////////////
// program entry
int main()
{
printf("Compile starting\n");
// build an expression tree:
// +
// / \
// + 4
// / \
// 2 3
struct Exp_t *exp = Exp_Sum_new (Exp_Sum_new(Exp_Int_new (2)
, Exp_Int_new (3))
, Exp_Int_new (4));
// print out this tree:
printf ("the expression is:\n");
Exp_print (exp);
// compile this tree to Stack machine instructions
compile (exp);
// print out the generated Stack instructons:
List_reverse_print (all);
printf("\nCompile finished\n");
return 0;
}