RPN简单实现

来源：互联网发布：中国软件培训编辑：程序博客网时间：2024/06/02 23:31
注意：结构实现的巧妙性
#include <iostream>using namespace std;#include <assert.h>#include <stack>enum Type//定义类型，操作数和操作符{    OP_SYMBOL,    OP_NUM,    ADD,    SUB,    MUL,    DIV,};struct Cell//单元{    Type _type;    int _value;};int CountRPN(Cell *rpn,size_t n)//注意传指针，RPN为数组{    assert(rpn);    stack<int> s;    for (size_t i = 0; i < n;++i)    {        if (rpn[i]._type == OP_SYMBOL)//注意取操作数时，顺序问题，如左右操作数不同可能影响结果，如：除法        {            int right = s.top();            s.pop();            int left = s.top();            s.pop();            switch (rpn[i]._value)             {            case ADD:                s.push(left + right);                break;            case SUB:                s.push(left - right);                break;            case MUL:                s.push(left * right);                break;            case DIV:                s.push(left / right);                break;            default:                assert(false);            }        }        else if (rpn[i]._type == OP_NUM)            s.push(rpn[i]._value);              }    return s.top();}//测试函数void Funtest(){    Cell RPN[] =     {        { OP_NUM, 12 },        { OP_NUM, 3 },        { OP_NUM, 4},        {OP_SYMBOL,ADD},        {OP_SYMBOL,MUL},        {OP_NUM,6},        {OP_SYMBOL,SUB},        { OP_NUM, 8 },        { OP_NUM,2 },        {OP_SYMBOL,DIV},        {OP_SYMBOL,ADD}    };    cout << CountRPN(RPN, sizeof(RPN) / sizeof(RPN[0]));}int main(){    Funtest();    getchar();    return 0;}
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