C语言综合实验3—计算器四则运算

题目要求：编程实现计算器程序，完成实数的加、减、乘、除运算。注意运算符优先级别。表达式要求采用中缀形式，例如：2.3+7.2*7
提示：表达式处理可参考“逆波兰表达式”范例。

完整程序如下：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAXSIZE 100
#define SYMBOLSIZE 50

//表达式运算符结构体
typedef struct oper_symbol
{
    char oper; //运算符
    unsigned int index;//运算符在表达式中的位置索引
}oper_symbol;

//表达式计算函数声明
double caculate(char * expression);
double get_data(char * str,int beg,int end);
int is_operator_symbol(char ch);
long int power(int base,int times);

int main()
{
    char *expression = NULL;
    double ret;
    expression = (char*)malloc(sizeof(char)*MAXSIZE);
    if(expression == NULL)
    {
        printf("内存分配失败.\n");
        return -1;
    }
    memset(expression,0,MAXSIZE*sizeof(char));
    while(1)
    {
        printf("请输入一个表达式: ");
        scanf("%s",expression);
        ret = caculate(expression);
        printf("表达式计算结果为: %lf\n",ret);
    }
    free(expression);
    system("pause");
    return 0;
}

//表达式计算函数声明
double caculate(char * expression)
{
    unsigned int i,j,k;
    unsigned int symbol_len = 0; //运算符栈的长度
    unsigned int datas_len = 0;  //操作数栈的长度
    unsigned int symbol_index_len=0; //运算符的个数
    double left_data,right_data; //左右操作数
    double temp,ret;

    double *datas = (double*)malloc(sizeof(double)*MAXSIZE);
    char *symbols = (char*)malloc(sizeof(char*)*SYMBOLSIZE);
    oper_symbol * symbols_index = (oper_symbol*)malloc(sizeof(oper_symbol)*SYMBOLSIZE);

    if(datas == NULL || symbols == NULL || symbols_index == NULL)
    {
        printf("内存分配失败.\n");
        return 0;
    }
    for(i=0;i<strlen(expression);i++)
    {
        if(is_operator_symbol(expression[i]))
        {
            symbols_index[symbol_index_len].oper = expression[i];
            symbols_index[symbol_index_len].index = i;
            symbol_index_len++;
        }
    }
    //根据运算符提取两边的数据进行计算
    for(j=0,i=0;j<symbol_index_len;j++)
    {
        //取左操作数
        if(datas_len == 0)
            left_data = get_data(expression,i,symbols_index[j].index);
        else
            left_data = datas[--datas_len];
        //取右操作数
        if( j+1 == symbol_index_len)
            right_data = get_data(expression,symbols_index[j].index+1,strlen(expression));
        else
            right_data = get_data(expression,symbols_index[j].index+1,symbols_index[j+1].index);
        //如果运算符是*和/，先计算结果在保存
        if(symbols_index[j].oper == '*' || symbols_index[j].oper == '/')
        {
            switch(symbols_index[j].oper)
            {
            case '*':
                temp = left_data * right_data;
                break;
            case '/':
                temp = left_data / right_data;
                break;
            }
            datas[datas_len++] = temp;
        }
        //否则将左右操作数都不错
        else
        {
            datas[datas_len++] = left_data;
            datas[datas_len++] = right_data;
            //记录+和-运算符的个数
            symbols[symbol_len++] = symbols_index[j].oper;
        }
    }
    //开始从后向前进行计算
    k = symbol_len;
    //直到操作符为空为止
    while(k)
    {
        //取出左右两个操作数
        left_data = datas[--datas_len];
        right_data = datas[--datas_len];
        k = symbol_len-1;
        switch(symbols[k])
        {
        case '+':
            temp = right_data + left_data;
            break;
        case '-':
            temp = left_data - right_data;
            break;
        }
        datas[datas_len++] = temp;
        symbol_len--;
    }
    ret = datas[0];
    free(datas);
    free(symbols);
    free(symbols_index);
    return ret;
}

//判断字符是否是运算符
int is_operator_symbol(char ch)
{
    if(ch == '+' || ch == '-' || ch == '*' || ch == '/')
        return 1;
    return 0;
}
//从字符串中提出去浮点数
double get_data(char * str,int beg,int end)
{
    int i,k;
    double ret = 0.0f;
    double integer = 0.0f;
    double decimal = 0.0f;
    double temp;
    i = beg;
    //取整数部分
    while(str[i] != '.' && i < end)
    {
        integer *= 10;
        integer += str[i] ^ 0x30;
        i++;
    }
    //取小数部分
    if(str[i] == '.')
    {
        i++;
        k=1;
        while(i<end)
        {
            temp =  str[i] ^ 0x30;
            temp /= power(10,k);
            decimal += temp;
            k++;
            i++;
        }
    }
    //整数部分加上小数部分
    ret = integer+decimal;
    return ret;
}

long int power(int base,int times)
{
    long ret = 1;
    int i;
    for(i=0;i<times;i++)
        ret *= base;
    return ret;
}

测试结果如下：