这是我的代码:
int main(int argc, char** argv) {
int sleeptime = atoi(argv[1]);
int sleepCount = atoi(argv[2]);
int sleepTimeRandom = 1 + (rand() % (int)(sleeptime));
int sleepCountRandom = 1 + (rand() % (int)(sleepCount));
sleepTimeRandom = sleepTimeRandom * 1000;
DWORD id = GetCurrentProcessId();
cout << "\n\nProcess with id " << id << " will sleep for " << sleepTimeRandom << " milliseconds\n";
cout << "\n\nProcess with id " << id << " will sleep " << sleepCountRandom << " times \n";
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当我打电话给exe时
sleeper 4 4
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我总是得到2000毫秒,4次......为什么?
rky*_*ser 12
您必须为随机数生成器播种.在这里看一个例子.
#include <stdio.h>
#include <stdlib.h>
int main ()
{
printf ("Random not seeded: %d\n", rand()%10);
return 0;
}
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Random not seeded: 3
Random not seeded: 3
Random not seeded: 3
Random not seeded: 3
Random not seeded: 3
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#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main ()
{
srand(time(NULL));
printf ("Random seeded: %d\n", rand()%10);
return 0;
}
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Random fast seeded: 7
Random fast seeded: 7
Random fast seeded: 7
Random fast seeded: 1
Random fast seeded: 1
Random fast seeded: 1
Random fast seeded: 5
Random fast seeded: 5
Random fast seeded: 5
Random fast seeded: 5
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如果您希望能够每秒多次调用您的实用程序,则必须使用不同的源来为您播种,否则您仍会得到一些重复的随机数.
这是一个使用纳秒而不是time()仅返回秒的示例.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main ()
{
struct timespec ts;
clock_gettime(CLOCK_MONOTONIC, &ts);
/* using nano-seconds instead of seconds */
srand((time_t)ts.tv_nsec);
printf ("Random fast seeded: %d\n", rand()%10);
return 0;
}
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您可以在此处看到,这些数字的分组与前一个示例不同.
Random fast seeded: 9
Random fast seeded: 6
Random fast seeded: 5
Random fast seeded: 6
Random fast seeded: 1
Random fast seeded: 1
Random fast seeded: 9
Random fast seeded: 4
Random fast seeded: 3
Random fast seeded: 2
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如果您对均匀分布的随机数感兴趣,请在下面看到user3003631的答案.如果你实际上在使用C++,那就是我建议做随机数的方法.关于这方面的更多信息.
如果您想要一种不同的方法,C++ 中也有随机效用...可能更高的质量
请原谅最少的解释
#include <iostream>
#include <algorithm>
#include <random>
int random_test_200()
{
std::random_device rd;
uniform_int_distribution<int> ud(1,9);
mt19937 mt(rd());
std::vector<int> v1;
for (int i = 0; i < 40; ++i)
{
auto x = ud(mt);
v1.push_back(x);
}
}
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