C++ 生成一个 4 位随机数
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C++ Generating a 4 digit random number
提问by Sudantha
I need to generate a 4 digit random number in C++
我需要在 C++ 中生成一个 4 位随机数
I used the following code
我使用了以下代码
#include<time.h>
int number;
number=rand()%1000;
srand ( time(NULL) );
but its doesn't gives a total random number
但它没有给出一个总的随机数
回答by Armen Tsirunyan
number = rand() % 9000 + 1000;
There are 9000 four-digit numbers, right? Starting from 1000 till 9999. rand()will return a random number from 0 to RAND_MAX. rand() % 9000will be from 0 to 8999 and rand() % 9000 + 1000;will be from 1000 to 9999 . In general when you want a random number from a to b inclusive the formula is
有 9000 个四位数，对吧？从 1000 到 9999.rand()将返回一个从 0 到 的随机数RAND_MAX。rand() % 9000将从 0 到 8999 ，rand() % 9000 + 1000;从 1000 到 9999 。一般来说，当你想要一个从 a 到 b 的随机数时，公式是
rand() % (b - a + 1) + a
Also note that srand()should be called only once and before any rand().
还要注意，srand()应该只调用一次并且在 any 之前调用rand()。
If you doconsider the numbers between 0 an 999 inclusive to be "four digit numbers", simply use rand() % 10000in that case. I don't consider them to be but I'm covering all bases, just in case.
如果您确实认为 0 到 999 之间的数字是“四位数字”，请rand() % 10000在这种情况下使用。我不认为它们是，但我涵盖了所有基础，以防万一。
HTH
HTH
回答by dalle
Remember to call srandonly once.
记住srand只调用一次。
Take a look at Boost.Random, if you don't like the distribution of rand.
如果您不喜欢.Random的分布，请查看Boost.Randomrand。
// produces randomness out of thin air
boost::mt19937 rng;
// distribution that maps to 1000..9999
boost::uniform_int<> fourdigits(1000,9999);
// glues randomness with mapping
boost::variate_generator<boost::mt19937&, boost::uniform_int<> > die(rng, fourdigits);
// simulate rolling a die
int x = die();
回答by log0
You must seed your random generator srandbeforecalling rand()
您必须srand在调用之前为您的随机生成器播种rand()
#include<time.h>
srand( time(NULL) );
int number = rand() % 10000; // number between 0 and 9999
