数据结构C语言编程—栈实现简单四则运算

水深无声 2023-03-13 02:58 9阅读 0赞

\#include<stdio.h>  
\#include<string.h>  
\#include<windows.h>  
\#include<malloc.h>  
\#define TRUE 1  
\#define FALSE 0  
\#define Elemtype int  
\#define elemtype char

char opt\[7\]\[7\]=  
\{  
\{’>’,’>’,’<’,’<’,’<’,’>’,’>’\},  
\{’>’,’>’,’<’,’<’,’<’,’>’,’>’\},  
\{’>’,’>’,’>’,’>’,’<’,’>’,’>’\},  
\{’>’,’>’,’>’,’>’,’<’,’>’,’>’\},  
\{’<’,’<’,’<’,’<’,’<’,’=’,’ ‘\},  
\{’>’,’>’,’>’,’>’,’ ‘,’>’,’>’\},  
\{’<’,’<’,’<’,’<’,’<’,’ ‘,’=’\},  
\};

char op\[7\]=\{’+’,’-’,’\*’,’/’,’(’,’)’,’\#’\};

typedef struct stack1  
\{  
Elemtype data;  
struct stack1 \*next;  
\}InitStack,\*LinkStack1;

typedef struct stack2  
\{  
elemtype optr;  
struct stack2 \*next;  
\}initStack,\*LinkStack2;

int Push\_dat(LinkStack1 top1,Elemtype x)  
\{  
LinkStack1 temp;  
temp=(LinkStack1)malloc(sizeof(InitStack));  
if(!temp)  
return FALSE;  
temp->data=x;  
temp->next=top1->next;  
top1->next=temp;  
return TRUE;  
\}

int Pop\_dat(LinkStack1 top1,Elemtype \*x)  
\{  
LinkStack1 temp;  
temp=top1->next;  
if(temp==NULL)  
return FALSE;  
top1->next=temp->next;  
\*x=temp->data;  
free(temp);  
return TRUE;  
\}

int GetTop\_dat(LinkStack1 top1)  
\{  
int x;  
LinkStack1 temp;  
temp=top1->next;  
if(temp==NULL)  
return FALSE;  
x=temp->data;  
return x;  
\}

int Push\_optr(LinkStack2 top2,elemtype ch)  
\{  
LinkStack2 temp;  
temp=(LinkStack2)malloc(sizeof(initStack));  
if(!temp)  
return FALSE;  
temp->optr=ch;  
temp->next=top2->next;  
top2->next=temp;  
return TRUE;  
\}

int Pop\_optr(LinkStack2 top2,elemtype \*ch)  
\{  
LinkStack2 temp;  
temp=top2->next;  
if(temp==NULL)  
return FALSE;  
top2->next=temp->next;  
\*ch=temp->optr;  
free(temp);  
return TRUE;  
\}

char GetTop\_optr(LinkStack2 top2)  
\{  
LinkStack2 temp;  
char c;  
temp=top2->next;  
if(temp==NULL)  
return FALSE;  
c=temp->optr;  
return c;  
\}

char Precede(char cr,char ch)  
\{  
int i,j,x1,x2;  
for(i=0;i<7;i++)  
\{  
if(crop\[i\])  
\{  
x1=i;  
break;  
\}  
\}  
for(j=0;j<7;j++)  
\{  
if(chop\[j\])  
\{  
x2=j;  
break;  
\}  
\}  
return opt\[x1\]\[x2\];  
\}

int IsOpr(char c)  
\{  
if(c==’+’||c==’-’||c==’\*’||c==’/’||c==’(’||c==’)’||c==’\#’)  
return 1;  
else  
return 0;  
\}

int Operate(int x,char opr,int y)  
\{  
int result;  
switch(opr)  
\{  
case’+’:  
result=x+y;  
break;  
case’-’:  
result=x-y;  
break;  
case’*’:  
result=x*y;  
break;  
case’/’:  
result=x/y;  
break;  
\}  
return result;  
\}

int ch\_dat(char ch)  
\{  
int x,i;  
i=(int)ch;  
switch(i)  
\{  
case 48:\{x=0;break;\}  
case 49:\{x=1;break;\}  
case 50:\{x=2;break;\}  
case 51:\{x=3;break;\}  
case 52:\{x=4;break;\}  
case 53:\{x=5;break;\}  
case 54:\{x=6;break;\}  
case 55:\{x=7;break;\}  
case 56:\{x=8;break;\}  
case 57:\{x=9;break;\}  
\}  
return x;  
\}

int EvaluateExpression()  
\{  
char ch,cr,theta,ch1;  
int result,a,b,tranvalue;  
InitStack OPND;  
initStack OPTR;  
LinkStack1 top1;  
LinkStack2 top2;  
top1=&OPND;  
top2=&OPTR;  
top1->next=NULL;  
top2->next=NULL;  
Push\_optr(&OPTR,’\#’);  
printf(“Please input an exprssion such as ‘3+7\*6\#’ :”);  
ch=getchar();  
while((ch!=’\#’)||(GetTop\_optr(&OPTR)!=’\#’))  
\{  
if(!IsOpr(ch))  
\{  
tranvalue=ch\_dat(ch);  
Push\_dat(&OPND,tranvalue);  
ch=getchar();  
\}  
else  
\{  
cr=GetTop\_optr(&OPTR);  
switch(Precede(cr,ch))  
\{  
case ‘<’:  
Push\_optr(&OPTR,ch);  
ch=getchar();  
break;  
case ‘=’:  
Pop\_optr(&OPTR,&ch);  
ch=getchar();  
break;  
case ‘>’:  
Pop\_optr(&OPTR,&theta);  
Pop\_dat(&OPND,&b);  
Pop\_dat(&OPND,&a);  
result=Operate(a,theta,b);  
Push\_dat(&OPND,result);  
break;  
\}  
\}  
\}  
printf(“The answer is :%d\\n”,GetTop\_dat(&OPND));  
\}

void main()  
\{  
char anw;  
printf("\\t\\tThe Simlpe Expreesion Evaluation\\n\\n");  
EvaluateExpression();  
getchar();  
while(1)  
\{ printf(“Do you want to continue?(Y/N) “);  
scanf(”%c”,&anw);  
printf("\\n");  
getchar();  
if(anw==‘Y’)  
\{  
EvaluateExpression();  
getchar();  
\}  
else if(anw==‘N’)  
break;  
\}  
\}