C语言从txt文本中读取多行用逗号分隔的数据

例如数据为：
0.5712,0.45356,0.74399
0.58775,0.4721,0.77793
0.51964,0.42228,0.69693
将每一列的数据保存在一个double型数组中。
解决方案：
1.利用fscanf()函数:

#include 
#include

int main()
{
    FILE *fp = fopen("C://Users//aoe//Desktop//2016.920//796-40//clo.txt", "r");
    if (fp == NULL)
    {
        printf("file open error\n");
        return -1;
    }

    double xx[3];
    double yy[3];
    double zz[3];

    for (int i = 0; i < 3; i++)
    {
        fscanf(fp, "%lf,%lf,%lf", &xx[i], &yy[i], &zz[i]);
        printf("%lf,%lf,%lf\n", xx[i], yy[i], zz[i]);
    }
    fclose(fp);
    system("pause");
    return 0;
}

参考：
http://blog.csdn.net/liangxanhai/article/details/8026496
http://blog.csdn.net/tomorrow_begin/article/details/38878109

2.利用fgets()函数

#include
#include
#include

double char2num(char *s);

int main()
{
    int i, j = 0, k;
    FILE*fp = fopen("C://Users//aoe//Desktop//2016.920//796-40//clo31.txt","r");
    if (fp == NULL)
    {
        printf("file open errror/n");
        return -1;
    }
    char buf[30];
    char str[10];

    double xx[176 * 144];
    double yy[176 * 144];
    double zz[176 * 144];

    memset(str, 0, sizeof(str));
    memset(buf, 0, sizeof(buf));

    int number = 0;
    double d = 0;
    for (int n = 0; n < 25344;n++)
    {
        fgets(buf, sizeof(buf), fp);
        for (i = 0; buf[i]; i++)
        {
            if (buf[i] != ',')
            {
                str[j] = buf[i];
                j++;
            }
            else if (buf[i] == ',')
            {
                printf("%s\t", str);
                d = char2num(str);
                number++;
                if (number == 1)
                    xx[n] = d;
                else if (number == 2)
                    yy[n] = d;
                memset(str, 0, sizeof(str));//void *memset(void *s,int c,size_t n) 总的作用：将已开辟内存空间 s 的首 n 个字节的值设为值 c。             
                j = 0;
            }
        }
        printf("%s", str);
        d = char2num(str);
        zz[n] = d;
        memset(str, 0, sizeof(str));
        j = 0;
        number = 0;
        memset(buf, 0, sizeof(buf));
    }
    fclose(fp);
    for (int i = 0; i < 25344; i++)
    {
        printf("%.5f\t", xx[i]);
        printf("%.5f\t", yy[i]);
        printf("%.5f\n", zz[i]);

    }
    system("pause");
    return 0;

}

double char2num(char *s) //将字符型数据转换成double型数据
{
    double d = 0;  
    int i, t = 0;  
    for (i = 0; s[i] != 0; i++)  
    { 
        if (s[i] == '.')   
        {
            t = 10;
        }
        else      
        {
            if (t == 0)    
            {
                d = d * 10 + (s[i] - '0'); 
            }
            else       
            { 
                d = d + (double)(s[i] - '0') / t;      
                t *= 10; 
            }
        } 
    }  
    return d; 
}

此方案会存在数据转换错误的情况，不建议采用，方法一比较简单。
参考：
http://www.cnblogs.com/xiaolongchase/archive/2011/10/22/2221326.html
http://blog.csdn.net/wuyu1125/article/details/7610652