C语言用栈编写求表达式的值

2025-04-02 00:24:50
推荐回答(3个)
回答1:

//表达式输入完了之后直接回车,就出结果了,跟平时输入字符串一样。

/**********************************************
算术表达式求值的算符优先级算法
利用栈来实现括号匹配和表达式求值
算法的详细说明,请查看清华大学出版社《数据结构》,严蔚敏&吴伟民著,3.3节
***********************************************/
#include
#include
#include
//存储空间初始分配量
#define STACK_INIT_SIZE 100
//存储空间分配增量
#define STACKINCREMENT 10
#define OK 0
#define ERROR 127

//定义一个顺序栈
typedef struct
{
int*base; //在栈构造之前和销毁之后,base的值为NULL
int*top; //栈顶指针
int stacksize; //当前已分配的存储空间,以元素为单位
}SqStack;
int InitStack(SqStack*S)
{
//构造一个空栈
S->base=(int*)malloc(STACK_INIT_SIZE*sizeof(SqStack));
if(NULL==S->base)
{ //内存分配失败
return ERROR ;
}
S->top=S->base ;
S->stacksize=STACK_INIT_SIZE ;

return OK ;
}
char GetTop(SqStack*S,char*element)
{
//若栈不空,取栈顶元素,用element返回
if(S->base==S->top)
{
return ERROR ;
}
*element=*(S->top-1);

return*element ;
}
int Push(SqStack*S,int element)
{
//插入元素element为新的栈顶元素
if((S->top-S->base)>S->stacksize)
{
//栈满,追加空间
S->base=(int*)realloc(S->base,(STACK_INIT_SIZE+STACKINCREMENT)*sizeof(SqStack));
if(NULL==S->base)
{
return ERROR ;
}
S->top=S->base+S->stacksize ;
S->stacksize+=STACKINCREMENT ;
}
*S->top++=element ;

return OK ;
}
int Pop(SqStack*S,int*element)
{
//若栈不为空,则删除栈顶元素,用element返回其值
if(S->top==S->base)
{
return ERROR ;
}
*element=*(--S->top);

return OK ;
}
int PopOPTR(SqStack*S,char*element)
{
if(S->top==S->base)
{
return ERROR ;
}
*element=*(--S->top);

return OK ;
}
//判断字符c是否在集合OP中
int InOP(char c,char OP[7])
{
for(int i=0;i<7;i++)
{
if(c==OP[i])
{
return OK ;
}
}

return ERROR ;
}
//判断运算符的优先级
int Compare(int a,int b)
{
if('+'==a)
{
switch(b)
{
case '+' :
return '>';
case '-' :
return '>' ;
case '*' :
return '<' ;
case '/' :
return '<' ;
case '(' :
return '<' ;
case ')' :
return '>' ;
case '\n' :
return '>' ;
}
}
if('-'==a)
{
switch(b)
{
case '+' :
return '>' ;
case '-' :
return '>' ;
case '*' :
return '<' ;
case '/' :
return '<' ;
case '(' :
return '<' ;
case ')' :
return '>' ;
case '\n' :
return '>' ;
}
}

if('*'==a)
{
switch(b)
{
case '+' :
return '>' ;
case '-' :
return '>' ;
case '*' :
return '>' ;
case '/' :
return '>' ;
case '(' :
return '<' ;
case ')' :
return '>' ;
case '\n' :
return '>' ;
}
}

if('/'==a)
{
switch(b)
{
case '+' :
return '>' ;
case '-' :
return '>' ;
case '*' :
return '>' ;
case '/' :
return '>' ;
case '(' :
return '<' ;
case ')' :
return '>' ;
case '\n' :
return '>' ;
}
}

if('('==a)
{
switch(b)
{
case '+' :
return '<' ;
case '-' :
return '<' ;
case '*' :
return '<' ;
case '/' :
return '<' ;
case '(' :
return '<' ;
case ')' :
return '=' ;
}
}

if(')'==a)
{
switch(b)
{
case '+' :
return '>' ;
case '-' :
return '>' ;
case '*' :
return '>' ;
case '/' :
return '>' ;
case ')' :
return '>' ;
case '\n' :
return '>' ;
}
}

if('\n'==a)
{
switch(b)
{
case '+' :
return '<' ;
case '-' :
return '<' ;
case '*' :
return '<' ;
case '/' :
return '<' ;
case '(' :
return '<' ;
case '\n' :
return '=' ;
}
}
return ERROR ;
}

//简单计算
int Calculate(int left,char oper,int right)
{
int result=0 ;
switch(oper)
{
case '+' :
result=left+right ;
break ;
case '-' :
result=left-right ;
break ;
case '*' :
result=left*right ;
break ;
case '/' :
result=left/right ;
break ;
}

return result ;
}

/**********************************************
算术表达式求值的算符优先级算法,设OPTR和OPND分别为运算符栈和运算数栈
OP为运算符集合
**********************************************/
int main()
{
SqStack OPTR,OPND ;
int element=0 ;
char OPTR_element ;
int leftNum,rightNum ;
char input ;
//获取输入
char OP[7]={'+','-','*','/','(',')','\n'};

InitStack(&OPTR);
Push(&OPTR,'\n');
InitStack(&OPND);
printf("请输入表达式\n");
input=getchar();
while('\n'!=input||'\n'!=GetTop(&OPTR,&OPTR_element))
{
int temp=0 ;
if(isdigit(input))
{
//如果是数字
ungetc(input,stdin);
//返回给输入流
scanf("%d",&temp);
Push(&OPND,temp);
//数字就进OPND栈
input=getchar();
continue ;
}

if(OK==InOP(input,OP))
{
GetTop(&OPTR,&OPTR_element);
switch(Compare(OPTR_element,input))
{
case '<' :
//栈顶元素优先级低
Push(&OPTR,input);
//运算符进OPTR栈
input=getchar();
break ;
case '=' :
//脱括号
PopOPTR(&OPTR,&OPTR_element);
input=getchar();
break ;
case '>' :
//退栈,并将运算结果入栈
PopOPTR(&OPTR,&OPTR_element);
Pop(&OPND,&rightNum);
Pop(&OPND,&leftNum);
Push(&OPND,Calculate(leftNum,OPTR_element,rightNum));
break ;
default :
printf("表达式括号不匹配\n");
return ERROR ;
}//switch
}//if

else
{
printf("表达式内有未知字符,即将退出\n");
return ERROR ;
}
}//while

int value ;
Pop(&OPND,&value);
printf("结果 = %d\n",value);

return OK ;
}//end

回答2:

你先这样改一下。

MAIN函数中定义的时候:

Stack_N OPN, *pOPN = &OPN;
Stack_T OPT, *pOPT = &OPT;

然后全部用这两个指针。

你那是内存错误,因为你根本就没有一个栈的实体,却一直用指针操作它。。。

回答3:

栈越界了吧

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