利用栈将简单中缀表达式转为后缀表达式的测试

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/* 通过使用栈将简单四则运算中缀表达式转换为后缀表达式 * '+', '-', '*', '/', '(', ')'共六种符号 * 原理:   1. 运算优先级规则是从左至右, 先乘除后加减, 有括号先算括号   2. 逐个扫描表达式, 将遇到的操作数都按原样输出, 碰到操作符时, 转3.   3. 扫描到的操作符与栈顶元素比较优先级, 栈顶元素的优先级大于等于扫描的操作符时,      栈顶操作符出栈输出, 继续重复该步骤, 直到优先级小于的条件, 将当前扫描的操作      符进栈.   4. 重复2和3步骤, 直到扫描结束, 依次将栈内操作符元素退栈到站空状态即止.   注意: 当遇到'('时, 优先级是最高的, 此时进栈, 遇到')'时, 需要退栈至'('.*/#include <stack>#include <stdio.h>#include <string.h>#include <stdlib.h>static bool is_operator(const char c);static int priority_compare(const char operator1, const char operator2);//中缀表达式转后缀表达式static void infix_express_to_suffix_express(const char *infix_express, int infix_len,                                            char *suffix_express, int suffix_len){    int i, j = 0;    std::stack<char> operator_stack;    int flag;    const char *p = infix_express;    for (i = 0; i<infix_len; ++i)    {        flag = 0;        if (!is_operator(p[i]))        {            suffix_express[j++] = p[i];        }        else        {            if (p[i] == ')')            {                while (!operator_stack.empty() && flag == 0)                {                    if ('(' == operator_stack.top())                    {                        flag = 1;                    }                    else                    {                        suffix_express[j++] = operator_stack.top();                    }                    operator_stack.pop();                }                if (flag == 0)                {                    printf("1. express invalid, exit. \n");                    exit(-1);                }                continue;            }            else            {                while (!operator_stack.empty())                {                    int priority_ret = -1;                    if ((priority_ret = priority_compare(operator_stack.top(), p[i])) == -1)                    {                        printf("2. invalid express, exit.\n");                        exit(-2);                    }                    if (priority_ret == 1)                    {                        suffix_express[j++] = operator_stack.top();                        operator_stack.pop();                    }                    else                        break;                }                operator_stack.push(p[i]);            }        }    }    while (!operator_stack.empty())    {        suffix_express[j++] = operator_stack.top();        operator_stack.pop();    }    suffix_express[j] = '\0';}static bool is_operator(const char c){    if (c == '+'        || c == '-'        || c == '*'        || c == '/'        || c == '('        || c == ')')    return true;    return false;}static int priority_compare(const char operator1, const char operator2){    switch (operator1)    {        case '+':        case '-':        {            switch (operator2)            {                case '+':                case '-':                    return 1;                break;                case '*':                case '/':                case '(':                    return 0;                break;                default:                    return -1;                break;            }        }        break;        case '*':        case '/':        {            switch (operator2)            {                case '+':                case '-':                case '*':                case '/':                    return 1;                break;                case '(':                    return 0;                break;                default:                    return -1;                break;            }        }        break;        case '(':        {            switch (operator2)            {                case '+':                case '-':                case '*':                case '/':                case '(':                    return 0;                default:                    return -1;                    break;            }        }        break;        default:            return -1;        break;    }    return -1;}int main(void){    const int MAX_EXPRESS_LENGTH = 256;    char express_infix[MAX_EXPRESS_LENGTH] = {'\0'};    char express_suffix[MAX_EXPRESS_LENGTH] = {'\0'};    printf("Input infix express('+', '-', '*', '/', '(', ')')\n");    scanf("%s", express_infix);    infix_express_to_suffix_express(express_infix, strlen(express_infix),                                    express_suffix, MAX_EXPRESS_LENGTH);    printf("suffix express=%s\n", express_suffix);    return 0;}
输入：a+b*c+(d*e+f)*g
输出：suffix express=abc*+de*f+g*+
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