请各位高手帮忙，写一个四则运算的算法，最好是Ｃ程序。例如可以根据字符串 1+2+3+4+5*6+((7+8)/5) 求出结果的程序或算法。急!

楼上的,你用的是什么出版社出版的书呀?把书名和出版社名写出来给大家呀!

___________________________________________

I'm Programer !

帖一个我n年前写的程序，基本上满足楼主的条件（在TC2.0里面通过）：

程序代码：

#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);
}

楼上的写的不错 正是最典型的栈操作

http://bbs.bc-cn.net/bbs/dispbbs.asp?boardID=39&ID=442&page=1

楼主若是想看计数器及四则混合运算方面的 就看上面的连接把