linux开发---20.【C】十六进制字符串转十进制

1.代码

#include "utils.h"

long long htoi(const char* s, int len)
{
    int i;
    long long n = 0;
    //现在保证s中是合法的并且都是小写的16进制字符串了,直接转就可以了
    for (i=0; i<len; i++)
    {
        if (s[i] > '9')
        {
            n = 16*n + (10+ s[i]-'a');
        }
        else
        {
            n = 16*n + s[i]-'0';
        }
    }
    return n;
}

int check_and_tran(char *d, char* s, int len)
{
    int i;
    //把去掉字符串开头的0x或0X，拷贝到d中
    if (s[0] == '0' && (s[1]=='x' || s[1]=='X')) //0x or 0X, prefix
        i=2;
    else
        i=0;
    strncpy(d, s+i, len-i);
    //printf("before d=%s\n", d);

    //判断d中是不是合法字符,并把其中的大写字母转小写
    for(i=0; i<strlen(d); i++)
    {
        // printf("%d=%c  ", i, d[i]);
        if( !((d[i]>='0' && d[i]<='9') || (d[i]>='a' && d[i]<='f') || (d[i]>='A' && d[i]<='F')) )
        {
            printf("not a valid hex string\n");
            return -1;
        }
        if(d[i]>='A' && d[i]<='F')
            d[i] += 'a' - 'A';
    }
    //printf("d=%s\n", d);
    return 0;
}

int main ( int argc, char *argv[] )
{
    int ret;
    char buf[1024];
    int len;
    long long num;
    if(argc < 2)
    {
        printf("./cal \n");
        return -1;
    }
    memset(buf, 0 , 1024);
    //1. 先检查传入的字符串是否是合法的16进制字符串
    //把字符串中的A-F转为a-f,转为小写字母
    ret = check_and_tran(buf, argv[1], strlen(argv[1]));
    if(ret < 0)
        return -1;

    //2. 将16进制字符串转为10进制的数
    num = htoi(buf, strlen(buf));

    //3.最后按照合适的顺序打印出来
    printf("0x%s=%lld\t", buf,num);
    if(num > 1024)
        printf("%lfK\t",(double)((double)num/1024));
    if(num > 1024*1024)
        printf("%lfM\t",(double)((double)num/1024/1024));
    if(num > 1024*1024*1024)
        printf("%lfG\t",(double)((double)num/1024/1024/1024));
    printf("\n\n");
    return EXIT_SUCCESS;
}

2. 代码打包
calculator.rar(下载后改改为calculator.tar.gz)
3. 实验结果

cong@msi:/work/test/calculator$ ./cal E4B00010
e4b00010=3836739600    3746816.015625K    3659.000015M    3.573242G

cong@msi:/work/test/calculator$ ./cal 1234
1234=4660    4.550781K