[英]How to cast 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
unsigned long
值,但散列函数采用unsigned char *
,如下面的实现中所示:
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? 有没有办法实现我的目标,也许是两者之间的演员?
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. 确保在
p
使用全部4个字节,或者在系统上使用unsigned long
的长度。
Technically you can achieve it with: 从技术上讲,你可以实现它:
unsigned long value = 58281;
djb2((unsigned char *) &value, sizeof(value));
Mind the usual pitfalls, however: 但要注意常见的陷阱:
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. sizeof(object) > (int) sizeof(object)
非常大的对象(如果适用于你的体系结构),请注意你可能会超出边界访问(未定义的行为)或仅你的对象的一部分哈希。 As other said, you can easily read an int
or any other object as a char
array : 正如其他人所说,您可以轻松地将
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 . 但请注意,存储在
short
或long
中的0xdead
将不会具有相同的哈希值 。
Also note that your hash function could be better unrolled using a Duff's device : 另请注意,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;
}
This shows a cast working. 这显示了演员阵容。 Note that in this case the "ABC" string will be null terminated but this may require more care in real world cases
请注意,在这种情况下,“ABC”字符串将被空终止,但这可能需要在现实世界中更加小心
#include <stdio.h>
int main() {
unsigned long x=0x414243; #0x414243 is ABC
unsigned char *s=(unsigned char *)&x;
printf("%s", s);
}
Since you've posted your code now, you'd want to use something similar to this: 由于您现在已经发布了代码,因此您需要使用与此类似的内容:
#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;
}
你应该使用ultoa_s转换它
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