这个程序不识别 负 号 那位能帮帮忙指点一下把这个程序改一下，能识别 负 号 怎样区别 负 号 与 减号 的区别？
我想了好久，就是想不明白怎么办~~~
而且要是 运算中有括号 怎么办？
是不是在用户输入数据的时候，如果系统遇到左括号，那么就记下这个括号所在数组的地址，依次类推，用if 嵌套来进行判断哪两个括号是一对的？请高手指点一下~~~~ 但是具体怎么实现呢~~~~
大家 高手 们 帮帮忙
#include<stdio.h>
#include<stdlib.h>
struct stack_node
{ int data;
struct stack_node *next;
};
typedef struct stack_node stack_list;
typedef stack_list *link;
link operator=NULL;
link operand=NULL;
link push(link stack, int value)
{ link new_node;
new_node=(link)malloc(sizeof(stack_list));
if(!new_node)
{
printf("Overflow!\n");
return NULL;
}
new_node->data=value;
new_node->next=stack;
stack=new_node;
return stack;
}
link pop(link stack, int *value)
{ link top;
if(stack!=NULL)
{
top=stack;
stack=stack->next;
*value=top->data;
free(top);
return stack;
}
else
*value=-1;
}
int empty(link stack)
{ if(stack==NULL)
return 1;
else
return 0;
}
int isoperator(char op)
{ switch(op)
{
case '+':
case '-':
case '*':
case '/': return 1;
default: return 0;
}
}
int priority(char op)
{ switch(op)
{
case ')': return 3;
case '*':
case '/': return 2;
case '+':
case '-': return 1;
case '(': return -1;
default: return 0;
}
}
int get_value(int op, int operand1, int operand2)
{ switch((char)op)
{
case '*': return(operand2*operand1);
case '/': return(operand2/operand1);
case '-': return(operand2-operand1);
case '+': return(operand2+operand1);
}
}
void main()
{ char exp[100];
int op=0;
int operand1=0;
int operand2=0;
int result=0;
int pos=0;
printf("Please input the expression:\n");
gets(exp);
while(exp[pos]!='\0'&&exp[pos]!='\n')
{
if(exp[pos]==' ')
{
pos++;
continue;
}
if(exp[pos]=='(') operator=push(operator,exp[pos]);
else if(exp[pos]==')')
{
operator=pop(operator,&op);
while(op!='(')
{
operand=pop(operand,&operand1);
operand=pop(operand,&operand2);
operand=push(operand,get_value(op,operand1,operand2));
operator=pop(operator,&op);
}
}
else if(isoperator(exp[pos]))
{
while(priority(exp[pos])<=priority(operator->data)&&!empty(operator))
{
operator=pop(operator, &op);
operand=pop(operand, &operand1);
operand=pop(operand, &operand2);
operand=push(operand,get_value(op,operand1,operand2));
}
operator=push(operator,exp[pos]);
}
else
operand=push(operand,exp[pos]-48);
pos++;
}
while(!empty(operator))
{
operator=pop(operator,&op);
operand=pop(operand,&operand1);
operand=pop(operand,&operand2);
operand=push(operand,get_value(op,operand1,operand2));
}
operand=pop(operand,&result);
printf("The result of expression [%s] is: %d\n",exp,result);
}
。。。。。。