问题描述
我有一个指向结构的指针数组。
我试图用这个问题的第一个答案来解决这个问题，但是我得到了一个句段错误：
尝试此代码后，我尝试使用此功能进行混洗：
static void shuffle（void * array，size_t n，size_t size）{
void * aux;
aux = malloc（大小）;
if（n> 1）{
size_t i;
for（i = 0; i< n-1; ++ i）{
size_t j = i + rand（）/（RAND_MAX /（n-i）+ 1） ;
memcpy（aux，array [j]，size）;
memcpy（array [j]，array [i]，size）;
memcpy（array [i]，aux，size）;
}
}
}
但是我遇到了以下错误：
警告：取消引用'void *'指针[默认启用]
错误：无效使用void表达式
我收到此警告和错误3次：每个memcpy（）一次。
编辑
我更改了代码，因为我试图使用新的而不是旧的。
除了@Deduplicator好的建议，还需要将 j 的偏移量缩放为 size 。
此代码还更改了 j 的计算。我不认为您所拥有的会很好。
#include< stdlib.h>
static void shuffle（void * array，size_t n，size_t size）{
//此if（）在功能上不是必需的，而是按照OP的样式保留
if（n> 1） {
char * carray = array;
void * aux;
aux = malloc（size）;
size_t i;
for（i = 1; i size_t j = rand（）％（i + 1）;
j * =大小；
memcpy（aux，& carray [j]，size）;
memcpy（& carray [j]，& carray [i * size]，size）;
memcpy（& carray [i * size]，aux，size）;
}
free（aux）;
}
}
阐明 size_t j = i + rand（）/（RAND_MAX /（n-i）+1）;
的弱点
将值 0 的 rand（）替换为 RAND_MAX 将生成样本分布。示例（i = 0，n = 30265，RAND_MAX = 2147483647）生成值 j = 0到30263 ，分别为70957次和 j = 30264 =41000。这是一个坏情况，可能不是最坏的情况。
使用 size_t j = i + r％ （ni）; 生成几乎统一结果。
示例（i = 0，n = 30265，RAND_MAX = 2147483647）生成值 j = 0至307 ，每个值70957次，而 j = 308至30264 = 70956次。存在补偿这种小的偏差的方法。许多SO帖子。
注意：这两种方法都不能解决 rand（）的继承弱点，但是当尝试使用 0 到 RAND_MAX
孔出现 size_t j = i + rand（）/（RAND_MAX /（n-i）+ 1）;
大约 n> sqrt（RAND_MAX）和 i = 0 ，将不会生成 n 附近的值。例如：（i = 0，n = 104643，RAND_MAX = 2147483647）不会生成值104638至104642。
I have an array of pointer to a struct.
I tried to use the first answer of this question to try to solve this problem, but I got a seg fault:
After try this code, I tryed to use this function to do the shuffling:
static void shuffle(void *array, size_t n, size_t size) {
void * aux;
aux = malloc (size);
if (n > 1) {
size_t i;
for (i = 0; i < n - 1; ++i) {
size_t j = i + rand() / (RAND_MAX / (n - i) + 1);
memcpy(aux, array[j], size);
memcpy(array[j], array[i], size);
memcpy(array[i], aux, size);
}
}
}
But i got the following error:
warning: dereferencing 'void *' pointer [enable by default] error: invalid use of void expression
I got this warning and error 3 times: one for each memcpy().
EDIT
I changed the code because I was trying to use this new one instead the old one.
In addition to @Deduplicator good advice, need to scale the j offset by size.
This code also changes the j calculation. I do not think what you have will work well.
#include <stdlib.h>
static void shuffle(void *array, size_t n, size_t size) {
// This if() is not needed functionally, but left per OP's style
if (n > 1) {
char *carray = array;
void * aux;
aux = malloc(size);
size_t i;
for (i = 1; i < n; ++i) {
size_t j = rand() % (i + 1);
j *= size;
memcpy(aux, &carray[j], size);
memcpy(&carray[j], &carray[i*size], size);
memcpy(&carray[i*size], aux, size);
}
free(aux);
}
}
Clarify the weakness of size_t j = i + rand() / (RAND_MAX / (n - i) + 1);
Substituting rand() with the values 0 to RAND_MAX in the formula will generate a sample distribution. Example (i=0, n=30265, RAND_MAX=2147483647) generates the values j = 0 to 30263, 70957 times each and j = 30264= 41000. This is a bad case and likely not the worst case.
Using size_t j = i + r%(n-i); generates a nearly uniform result.
Example (i=0, n=30265, RAND_MAX=2147483647) generates the values j = 0 to 307, 70957 times each and j = 308 to 30264= 70956 times. Methods exist to compensate for this small bias. Many SO postings.
Note: Inherit weaknesses in a poor rand() are not fixed by either of these methods, but method selection can make things noticeably worse when trying to use a sub-range of 0 to RAND_MAX
[Edit]
Holes appear with size_t j = i + rand() / (RAND_MAX / (n - i) + 1);
Somewhere about n > sqrt(RAND_MAX) and i=0, values near n will not be generated. Example: (i=0, n=104643, RAND_MAX=2147483647) does not generate values 104638 to 104642.
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