如何将八进制转换为二进制、十进制、十六进制

1.八进制 –> 二进制
我这里采用了一种简单又高效的办法.
先把000 - 111 的二进制用二维数组存起来.
例如:八进制25–> 5对应101 2对应010 这时候只要把八进制分解为一位一位的,然后在数组中找对应的二进制即可,无论多大的八进制数,都可以很快的分解并找到对应的二进制数.

    char a[9][5] = { "000 ","001 ","010 ","011 ","100 ","101 ","110 ","111 " };    int temp[20] = { 0 }; //存放每一位八进制数,例如:25 --> temp[0] = 5 ,temp[1] = 2    int iLen = 0, i = 0, j = 0;    while (oct) //分解八进制数    {        temp[i++] = oct % 10;        if (temp[i - 1] == 8 || temp[i - 1] == 9)            return -1; //八进制数只能是0-7        oct /= 10;    }    do    {        --i;        //例如:temp[1] = 2 --> a[2][] = "010 ",temp[0] = 5 --> a[5][] = "101 "        while (j < 4) //    --> bin = " 010 101"             bin[iLen++] = a[temp[i]][j++];        j = 0;    } while (i);    return 0;} 

注意:这个函数最大支持19位8进制数的转换
2.八进制 –> 十进制

这个也是权值相加即可.

long long OctToDec(long long oct){    long long dec = 0;    int p = 0, t = 0;    while (oct)    {        t = oct % 10;        dec += powl(8, p++)*t;        if (t == 8 || t == 9)            return -1;        oct = oct / 10;    }    return dec;}

注意:这个函数最大支持19位8进制数的转换

3.八进制 –> 16进制

只能通过间接转换了。

int OctToHex(long long oct, char* hex){    long long dec = 0;    char a[17] = "0123456789ABCDEF";    char temp[30] = " ";    int p = 0, t = 0, i = 0, j = 0;    while (oct)               //八进制转换为十进制    {        t = oct % 10;        dec += powl(8, p++)*t;        if (t == 8 || t == 9)            return -1;        oct = oct / 10;    }    while (dec)               //十进制转换为十六进制     {        temp[i++] = a[dec % 16];        dec = dec >> 4;    }    while (i)        hex[j++] = temp[--i]; //反转    return 0;}