C++中从-9到9的随机数
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/7887941/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
random number from -9 to 9 in C++
提问by Danny
just wondering, if I have the following code:
只是想知道，如果我有以下代码：
int randomNum = rand() % 18 + (-9);
will this create a random number from -9 to 9?
这会创建一个从 -9 到 9 的随机数吗？
回答by NPE
No, it won't. You're looking for:
不，不会。您正在寻找：
int randomNum = rand() % 19 + (-9);
There are 19 distinct integers between -9 and +9 (including both), but rand() % 18only gives 18 possibilities. This is why you need to use rand() % 19.
-9 和 +9 之间有 19 个不同的整数（包括两者），但rand() % 18只给出了 18 种可能性。这就是为什么您需要使用rand() % 19.
回答by Davide Aversa
Your code returns number between (0-9 and 17-9) = (-9 and 8).
您的代码返回 (0-9 和 17-9) = (-9 和 8) 之间的数字。
For your information
供您参考
rand() % N;
returns number between 0 and N-1 :)
返回 0 到 N-1 之间的数字:)
The right code is
正确的代码是
rand() % 19 + (-9);
回答by KillianDS
Do not forget the new C++11 pseudo-random functionality, could be an option if your compiler already supports it.
不要忘记新的C++11 伪随机功能，如果您的编译器已经支持它，它可能是一个选项。
Pseudo-code:
伪代码：
std::mt19937 gen(someSeed);
std::uniform_int_distribution<int> dis(-9, 9);
int myNumber = dis(gen)
回答by nurettin
You are right in that there are 18 counting numbers between -9 and 9 (inclusive).
你是对的，在 -9 和 9（含）之间有 18 个计数数字。
But the computer uses integers (the Z set) which includes zero, which makes it 19 numbers.
但是计算机使用包括零的整数（Z 集），这使它成为 19 个数字。
Minimum ratio you get from rand() over RAND_MAX is 0, so you need to subtract 9 to get to -9.
您从 rand() 获得的最小比率与 RAND_MAX 为 0，因此您需要减去 9 才能得到 -9。
The information below is deprecated. It is not in manpages aymore. I also recommend using modernC++for this task.
以下信息已弃用。它不在联机帮助页中。我还建议使用现代C++来完成这项任务。
Also, manpage for the rand functionquotes:
"If you want to generate a random integer between 1 and 10, you should always do it by using high-order bits, as in
j = 1 + (int) (10.0 * (rand() / (RAND_MAX + 1.0)));and never by anything resembling
j = 1 + (rand() % 10);(which uses lower-order bits)."
“如果你想生成一个 1 到 10 之间的随机整数，你应该总是使用高位来做，如
j = 1 + (int) (10.0 * (rand() / (RAND_MAX + 1.0)));从来没有类似的东西
j = 1 + (rand() % 10);（使用低位）。”
So in your case this would be:
所以在你的情况下，这将是：
int n= -9+ int((2* 9+ 1)* 1.* rand()/ (RAND_MAX+ 1.));
回答by Petruza
Anytime you have doubts, you can run a loop that gets 100 million random numbers with your original algorithm, get the lowest and highest values and see what happens.
任何时候你有疑问，你都可以运行一个循环，用你的原始算法得到 1 亿个随机数，得到最低和最高值，看看会发生什么。
