字符串转浮点数

来源：互联网发布：windows查看隐藏文件编辑：程序博客网时间：2024/05/18 01:26
自己实现的字符串转换浮点数的代码
模电期中考试无聊写的- -
先记录一下，以后有时间在优化。
精度之类的等等问题也暂时没有考虑
// =====================================================================================////       Filename:  my_atoi.c////    Description:  ////        Version:  1.0//        Created:  2014年11月14日 19时19分48秒//       Revision:  none//       Compiler:  g++////         Author:  wangweihao (wangweihao), 578867817@qq.com//        Company:  none//// =====================================================================================/* * 实现字符转转浮点函数 * 浮点数可能有普通浮点数和科学计数法的浮点数 * 情况比较多 * 举个例子 * 3.14  0.314  314   3.14e10(等价于3.14e+10)    3.14e-10    */#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>double my_atoi(char *s){    int i, length, newlength;    int flag = 0, flag2 = 0;           //标记'.' , 和'e'    int flagIndex = 0, flagIndex2 = 0;    char symbol;    double sum = 0;                    //计算最后的总和    double sum1 = 0, sum2 = 0, sum3 = 0;    double ten1 = 1, ten2 = 0.1, ten3 = 1;    length = strlen(s);    for(i = 0; i < length; ++i)    {        if(isdigit(s[i]))            ;        else if(ispunct(s[i]))        {            if(s[i] == '.')            {                flagIndex = i;                ++flag;            }        }        else if(isalpha(s[i]))        {            if(s[i] == 'e')            {                ++flag2;                flagIndex2 = i;            }            if(i < length)            {                symbol = s[i+1];   //存储符号            }        }    }        if(flag > 1 || flag2 > 1)    {        printf("error input\n");        exit(-1);    }    //printf("flag = %d, flagIndex = %d, flag2 = %d, flagIndex2 = %d\nsymbol = %c\n", flag, flagIndex, flag2, flagIndex2, symbol);    if(flag2 != 0)        length = flagIndex2;    if(flag == 0 && flag2 == 0)    {        for(i = length-1; i >= 0; --i)        {            sum += (s[i]-48)*ten1;            ten1 *= 10;        }        return sum;    }    for(i = flagIndex-1; i >= 0; --i)    {        sum1 += (s[i]-48)*ten1;        ten1 *= 10;    }    for(i = flagIndex+1; i < length; ++i)    {        sum2 += (s[i]-48)*ten2;        ten2 /= 10;    }    length = strlen(s);    for(i = length-1; isdigit(s[i]); --i)    {        sum3 += (s[i]-48)*ten3;        ten3 *= 10;    }    if(flag2 != 0)    {        if(symbol == '-')        {            sum3 = -sum3;            sum = (sum1 + sum2)* (pow(10, sum3));            return sum;        }        else            sum = (sum1 + sum2)* (pow(10, sum3));        return sum;    }        sum = sum1 + sum2;    return sum;}int main(int argc, char *argv[]){    char *p1 = "3.14";    char *p2 = "0.314";    char *p3 = "314";    char *p4 = "3.14e3";    char *p5 = "3.14e+3";    char *p6 = "3.14e-3";    char a[100];    printf("p1:%lf, p2:%lf, p3:%lf, p4:%lf, p5:%lf, p6:%lf\n", my_atoi(p1), my_atoi(p2), my_atoi(p3), my_atoi(p4), my_atoi(p5), my_atoi(p6));    scanf("%s", a);    printf("my_atio(%s): %lf\n", a, my_atoi(a));    return 0;
0 0