有关栈API详情參看我的还有一篇博文：

遍历后缀表达式中的数字和符号

对于数字：进栈

对于符号：

从栈中弹出右操作数

从栈中弹出左操作数

依据符号进行运算

将运算结果压入栈中

遍历结束：栈中的唯一数字为计算结果

#include 
     
      #include "LinkStack.h"int isNumber3(char c){	return ('0' <= c) && (c <= '9');}int isOperator3(char c){	return (c == '+') || (c == '-') || (c == '*') || (c == '/');}int value(char c){	return (c - '0');}int express(int left, int right, char op){	int ret = 0;	switch (op)	{	case '+':		ret = left + right;		break;	case '-':		ret = left - right;		break;	case '*':		ret = left * right;		break;	case '/':		ret = left / right;		break;	default:		break;	}	return ret;}int compute(const char* exp){	LinkStack* stack = LinkStack_Create();	int ret = 0;	int i = 0;	while (exp[i] != '\0')	{		if (isNumber3(exp[i]))		{			LinkStack_Push(stack, (void*)value(exp[i]));		}		else if (isOperator3(exp[i]))		{			int right = (int)LinkStack_Pop(stack);			int left = (int)LinkStack_Pop(stack);			int result = express(left, right, exp[i]);			LinkStack_Push(stack, (void*)result);		}		else		{			printf("Invalid expression!");			break;		}		i++;	}	if ((LinkStack_Size(stack) == 1) && (exp[i] == '\0'))	{		ret = (int)LinkStack_Pop(stack);	}	else	{		printf("Invalid expression!");	}	LinkStack_Destroy(stack);	return ret;}int main(){	printf("8 + (3 - 1) * 5  = %d\n", compute("831-5*+"));	return 0;}
     

project文件详情：