今天幫同學看程式碼，她想實現把16進位制的數轉換成2進位制顯示，百度了一下，程式碼是不少，但是都太長了，想起C和指標裡面有似曾相識的東西，就自己寫了一下，還是遇到了一些問題。

1. 沒有注意到字串儲存的時候後面會自動加一個\0，因此想存放一個長度為n的字串，需要開闢n + 1 個位元組的空間，最後一個位元組存放\0.

2. 迴圈過程中注意迴圈變數每次迭代過程中都要想終止條件靠近，避免死迴圈.

3.利用陣列下標的控制來模擬棧

廢話太多了，下面給出原始碼：

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


void
HextoTwo(int num)
{
    int res;
    int i = 0;
    char buf[BUFSIZ][5] = {"0000"};
    char reference[16][5] = {"0000","0001","0010","0011",\
                        "0100","0101","0110","0111",\
                        "1000","1001","1010","1011",\
                        "1100","1101","1110","1111"};

    while(num / 16 !=  0)
    {
        res = num % 16
 
;
        strcpy(buf[i++], reference[res]);
        num = num / 16;
    }

    res = num % 16;
    strcpy(buf[i++], reference[res]);

    while(i > 0)
        printf("%s ", buf[--i]);
}

int main()
{
    int num = 0xfa;
    HextoTwo(num);
    return EXIT_SUCCESS;
}

下面給出遞迴解法，話說天才用遞迴，然而本人很菜…….

void
 

HextoTwo(int num)
{
    int remainder;
    char buf[16][5] = {"0000","0001","0010","0011",\
                        "0100","0101","0110","0111",\
                        "1000","1001","1010","1011",\
                        "1100","1101","1110","1111"};
    if(0 == num)
        return;

    remainder = num % 16;
    HextoTwo(num >> 4);
    printf("%s ", buf[remainder]);
}

就這麼簡單，還是要多寫程式碼多思考，演算法沒有形成之前不要動手。