如何将long unsigned to unsigned char*?

I am trying to hash an unsigned long value, but the hash function takes an unsigned char *, as seen in the implementation below:

我试图散列一个未签名的长值，但哈希函数采用无符号的char *，如下面的实现所示:

unsigned long djb2(unsigned char *key, int n)
{
    unsigned long hash = 5381;
    int i = 0;
    while (i < n-8) {
        hash = hash * 33 + key[i++];
        hash = hash * 33 + key[i++];
        hash = hash * 33 + key[i++];
        hash = hash * 33 + key[i++];
        hash = hash * 33 + key[i++];
        hash = hash * 33 + key[i++];
        hash = hash * 33 + key[i++];
        hash = hash * 33 + key[i++];
    }
    while (i < n)
        hash = hash * 33 + key[i++];
    return hash;
}

Is there a way I can achieve my goal, perhaps with a cast between the two?

有没有一种方法可以让我实现我的目标，也许是在两者之间的转换?

6 个解决方案

#1

unsigned long x;

unsigned char * p = (unsigned char*)&x;

Make sure you use all 4 bytes through the p, or whatever is the length of unsigned long on your system.

确保您使用了所有4字节的p，或任何长度的无符号长在您的系统上。

#2

Technically you can achieve it with:

技术上你可以做到:

unsigned long value = 58281;
djb2((unsigned char *) &value, sizeof(value));

Mind the usual pitfalls, however:

不过要注意常见的陷阱:

• The hash function in question was originally meant for strings (hence the prototype), so make sure it fits your needs (# of collisions, avalanching, etc.)
• 问题中的哈希函数最初是用于字符串(因此原型)，所以要确保它符合您的需求(冲突的#，雪崩，等等)。
• If at some point you want to hash very large objects for which sizeof(object) > (int) sizeof(object) (if applicable on your architecture(s)), note you might get out of bounds accesses (undefined behaviour) or only part of your object hashed.
• 如果在某个点上，您想要散列非常大的对象，以便为其sizeof(object) > (int) sizeof(object)(如果适用于您的体系结构)，请注意，您可能会跳出边界访问(未定义的行为)，或者是您的对象散列的一部分。

#3

As other said, you can easily read an int or any other object as a char array :

正如其他人所说，您可以很容易地将int或任何其他对象读取为char数组:

unsigned char value = 0xde;
unsigned short value = 0xdead;
unsigned long value = 0xdeadbeef;
double value = 1./3;

djb2((unsigned char*)&value, sizeof value);

But note that 0xdead stored in a short or a long won't have the same hash.

但是请注意，在短时间或长时间内存储的0xdead不会有相同的散列。

Also note that your hash function could be better unrolled using a Duff's device :

还要注意，使用Duff的设备可以更好地展开哈希函数:

unsigned long djb2(unsigned char *k, int size)
{
    unsigned long h = 5381;
    int i = 0;
    switch(size % 8) {
      case 0: while(i < size) { 
                  h = h*33 + k[i++];
      case 7:     h = h*33 + k[i++];
      case 6:     h = h*33 + k[i++];
      case 5:     h = h*33 + k[i++];
      case 4:     h = h*33 + k[i++];
      case 3:     h = h*33 + k[i++];
      case 2:     h = h*33 + k[i++];
      case 1:     h = h*33 + k[i++];
              }
    }
    return h;
}

#4

This shows a cast working. Note that in this case the "ABC" string will be null terminated but this may require more care in real world cases

这显示了一个演员的工作。请注意，在这种情况下，“ABC”字符串将被终止，但这可能需要在现实世界中更多的关注。

#include <stdio.h>

int main() {
    unsigned long x=0x414243;  #0x414243 is ABC
    unsigned char *s=(unsigned char *)&x;
    printf("%s", s);
}

#5

Since you've posted your code now, you'd want to use something similar to this:

既然你现在已经发布了你的代码，你会想要使用类似的东西:

#include <stdio.h>


int main() {
    unsigned long result, x = 0xdeadbeef;
    x = convert_endian(x);

    result = djb2((unsigned char*)&x, sizeof(x));
    do_something(result);
    return 0;
}

#6

You should convert it using ultoa_s

你应该使用ultoa_s转换它。