生成用C独一无二的随机数 [英] Generating unique random numbers in C
问题描述
我试图产生10个独特的随机数在C.我有一个数组 numout [] 10号，但这到达段错误一段时间后。
芹苴code是：
INT I，J，numout [10]，randnum;无效的主要（）
{
函数srand（时间（NULL））;
对于（I = 0; I＆小于10;我+ +）
{
numout [I] =生成（）;
的printf（％d个，numout [I]）;
fflush（标准输出）;
睡眠（1）;
的printf（\\ B）;
}
}
INT产生（）
{
randnum = 1 +（INT）（RAND（）* mul_val /（RAND_MAX + 1.0））;
为（J = 0; J＆LT; I; J ++）
{
如果（randnum == 0 || randnum == numout [J]。）
{
randnum =生成（）;
}
}
返回（randnum）;
}
扔了code了，伤势严重。你需要一个洗牌的算法，不是一块code，对于重复检查旧的价值观。做你的方式最终会需要更长的时间和更长的池耗尽。混洗算法的优点在于，它不会降低为池变得更小。
下面是一张code我在回答不同的问题使用。它保持号码列表，并且当它返回一个随机您的话，其从列表中删除，递减为下一个随机选择的计
的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆stdlib.h中GT;＃定义ERR_NO_NUM -1
＃定义ERR_NO_MEM -2INT myRandom（INT大小）{
INT I，N;
静态INT numNums = 0;
静态INT * numArr = NULL; //与特定大小初始化。 如果（大小＆GT; = 0）{
如果（numArr！= NULL）
免费（numArr）;
如果（（numArr =的malloc（sizeof的（INT）*尺寸））== NULL）
返回ERR_NO_MEM;
对于（i = 0; I＆LT;大小;我++）
numArr [我] =我;
numNums =大小;
} //错误，如果没有号码池离开了。 如果（numNums == 0）
返回ERR_NO_NUM; //从池中获取随机数，并将其取出（RND在这
//情况下返回0和numNums-1含）之间的一个数字。 N =兰特（）％numNums;
I = numArr [N];
numArr [η] = numArr [numNums-1];
numNums--;
如果（numNums == 0）{
免费（numArr）;
numArr = 0;
} 返回我;
}诠释主要（无效）{
INT I; 函数srand（时间（NULL））;
I = myRandom（20）;
而（ⅰ＆GT; = 0）{
的printf（数=％3D \\ n，I）;
I = myRandom（-1）;
}
的printf（最终=％3D \\ n，I）;
返回0;
}
一个示例输出显示它在行动：
数= 19
数= 10
数= 2
数= 15
数= 0
数= 6
数= 1
总数= 3
数= 17
数= 14
数= 12
数= 18
数= 4
数= 9
数= 7
数= 8
数= 16
数= 5
数= 11
数= 13
最终= -1
用非负池大小调用它和它建立一个新的序列并返回第一个随机值。在此之后，你可以用 1 调用它，它会从池中获取下一个随机的，唯一编号。当池耗尽，它会返回-1。
借助包含此code等答案有一个版本能够保持多个泳池，以及如果你希望能够在线程code使用此功能。
I tried generating 10 unique random numbers in C. I have an array numout[] for 10 numbers but this gets to "segmentation fault" after some time.
Tho code is:
int i,j,numout[10],randnum;
void main()
{
srand(time(NULL));
for(i=0;i<10;i++)
{
numout[i]=generate();
printf("%d",numout[i]);
fflush(stdout);
sleep(1);
printf("\b");
}
}
int generate()
{
randnum=1+(int)(rand()*mul_val/(RAND_MAX+1.0));
for(j=0;j<i;j++)
{
if(randnum==0 || randnum==numout[j])
{
randnum=generate();
}
}
return(randnum);
}
Throw that code away, seriously. You need a shuffling algorithm, not a piece of code that checks older values for duplicates. Doing it your way will end up taking longer and longer as your pool runs out. The advantage of a shuffling algorithm is that it doesn't degrade as the pool becomes smaller.
Here's a piece of code I used in answering a different question. It maintains a list of numbers and, when it returns a random one to you, it removes it from the list and decrements the count for the next random selection.
#include <stdio.h>
#include <stdlib.h>
#define ERR_NO_NUM -1
#define ERR_NO_MEM -2
int myRandom (int size) {
int i, n;
static int numNums = 0;
static int *numArr = NULL;
// Initialize with a specific size.
if (size >= 0) {
if (numArr != NULL)
free (numArr);
if ((numArr = malloc (sizeof(int) * size)) == NULL)
return ERR_NO_MEM;
for (i = 0; i < size; i++)
numArr[i] = i;
numNums = size;
}
// Error if no numbers left in pool.
if (numNums == 0)
return ERR_NO_NUM;
// Get random number from pool and remove it (rnd in this
// case returns a number between 0 and numNums-1 inclusive).
n = rand() % numNums;
i = numArr[n];
numArr[n] = numArr[numNums-1];
numNums--;
if (numNums == 0) {
free (numArr);
numArr = 0;
}
return i;
}
int main (void) {
int i;
srand (time (NULL));
i = myRandom (20);
while (i >= 0) {
printf ("Number = %3d\n", i);
i = myRandom (-1);
}
printf ("Final = %3d\n", i);
return 0;
}
A sample output shows it in action:
Number = 19
Number = 10
Number = 2
Number = 15
Number = 0
Number = 6
Number = 1
Number = 3
Number = 17
Number = 14
Number = 12
Number = 18
Number = 4
Number = 9
Number = 7
Number = 8
Number = 16
Number = 5
Number = 11
Number = 13
Final = -1
Call it with a non-negative pool size and it sets up a new sequence and returns the first random value. Following that, you can call it with -1 and it will get the next random, unique number from the pool. When the pool is exhausted, it will return -1.
The other answer that contained this code has a version that can maintain multiple pools as well if you want to be able to use this function in threaded code.
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