编写程序对表达式求值C语言

编写程序对表达式求值C语言

#include "stdio.h"
#include "malloc.h"

#define TRUE 1
#define FALSE 0
#define OK 1
#define ERROR 0
#define INFEASIBLE -1
#define OVERFLOW -2

typedef int Status;
#define STACK_INIT_SIZE 100
#define STACKINCREMENT 10

typedef struct{
SElemType *base;
SElemType *top;
int stacksize;
}SqStack;
//构造一个空栈
Status InitStack(SqStack *S){
S->base = (SElemType*)malloc(STACK_INIT_SIZE*sizeof(SElemType));

if(!S->base)
exit(OVERFLOW);
S->top=S->base;
S->stacksize=STACK_INIT_SIZE;
return OK;
}
//判断是否为空栈
Status StackEmpty(SqStack S){
if(S.top == S.base)
return TRUE;
else
return FALSE;
}
//用e返回S的顶元素
Status GetTop(SqStack S, SElemType *e){
if(S.top == S.base)
return ERROR;
*e = *(S.top-1);
return OK;
}
//插入e为新的顶元素
Status Push(SqStack *S, SElemType e){
if((S->top - S->base) >= S->stacksize){
S->base = (
SElemType*)realloc(S->base,
(S->stacksize+STACKINCREMENT)*sizeof(SElemType)
);
if(!S->base)
exit(OVERFLOW);
S->top = S->base +S->stacksize;
S->stacksize += STACKINCREMENT;
}
*(S->top)=e;
S->top++;
return OK;
}
//删除S的顶元素，并用e返回其值
Status Pop(SqStack *S, SElemType *e){
if(S->top == S->base)
return ERROR;
S->top--;
*e = *(S->top);
return OK;
}
//从栈底到栈顶依次对S的每个元素调用函数Visit（），一旦失败操作无效
Status ListTraverse(SqStack S,Status (*visit)(SElemType)){
SElemType *p;
p=S.base;
for(p=S.base;p (*visit)(*p);

return OK;
}
//输出元素e
Status output(SElemType e){
printf("%d ",e);

return OK;
}

实现表达式求值的代码：

typedef int SElemType;
#include
#include "stack_s.c"

Status in(char c,char op[]){
char *p;
p=op;
while(*p != '\0'){
if(c == *p)
return TRUE;
p++;
}
return FALSE;
}

char Precede(char a, char b){
int i,j;
char pre[][7]={

{'>','>','<','<','<','>','>'},
{'>','>','<','<','<','>','>'},
{'>','>','>','>','<','>','>'},
{'>','>','>','>','<','>','>'},
{'<','<','<','<','<','=','0'},
{'>','>','>','>','0','>','>'},
{'<','<','<','<','<','0','='}};
switch(a){
case '+': i=0; break;
case '-': i=1; break;
case '*': i=2; break;
case '/': i=3; break;
case '(': i=4; break;
case ')': i=5; break;
case '#': i=6; break;
}
switch(b){
case '+': j=0; break;
case '-': j=1; break;
case '*': j=2; break;
case '/': j=3; break;
case '(': j=4; break;
case ')': j=5; break;
case '#': j=6; break;
}
return pre[i][j];
}

int Operate(int a, char theta, int b){
int i,j,result;
i=a;
j=b;

switch(theta) {
case '+': result = i + j; break;
case '-': result = i - j; break;
case '*': result = i * j; break;
case '/': result = i / j; break;
}
return result;
}

int getNext(int *n){
char c;
*n=0;
while((c=getchar())==' ');
if(!isdigit(c)){
*n=c;
return 1;
}
do {
*n=*n*10+(c-'0');
c=getchar();
} while(isdigit(c));
ungetc(c,stdin);
return 0;
}

int EvaluateExpression(){

int n;
int flag;
int c;
char x,theta;
int a,b;

char OP[]="+-*/()#";
SqStack OPTR;
SqStack OPND;

InitStack(&OPTR);
Push(&OPTR,'#');
InitStack(&OPND);
flag=getNext(&c);

GetTop(OPTR, &x);
while(c!='#' || x != '#')
{
if(flag == 0)
{
Push(&OPND,c);
flag = getNext(&c);
} else
{
GetTop(OPTR, &x);
switch(Precede(x, c))
{
case '<'://栈顶元素优先级低
Push(&OPTR,c);
flag = getNext(&c);
break;
case '='://脱括号并接受下一字符
Pop(&OPTR,&x);
flag = getNext(&c);
break;
case '>'：// 退栈并将运算结果入栈
Pop(&OPTR, &theta);
Pop(&OPND,&b);
Pop(&OPND,&a);
Push(&OPND, Operate(a, theta, b));
break;
}
}
GetTop(OPTR, &x);
}
GetTop(OPND, &c);
return c;
}

void main(){
int c;
printf("Please input one expression:");
c=EvaluateExpression();
printf("Result=%d\n",c);
getch();
}

复合运算。这个我会，可以帮你写。