How to generate random numbers between 2 values, inclusive? [duplicate] - c++

Question

This question already has answers here:
Generating a random integer from a range
(14 answers)
Closed 6 years ago.
I need help figuring out how to generate a set amount of random numbers between two user-inputted values, inclusively. Just so you know, I have searched for this but am not finding what I have just stated. Here is my code:
#include <iostream>
#include <ctime>
#include <cstdlib>
using namespace std;
int main()
{
int userBeg, userEnd, outPut;
cout << "Enter a start value: ";
cin >> userBeg;
cout << "Enter an end value: ";
cin >> userEnd;
srand(time(NULL)); //generates random seed val
for (int i = 0; i < 5; i++) {
//prints number between user input, inclusive
outPut = rand()%((userEnd - userBeg) + 1);
cout << outPut << " ";
}//end for
return 0;
}//end main
I'm confused with the output I get for the following ranges:
1-100 yield output numbers which fall in-between, but not including, the boundaries such as 50, 97, 24, 59, 22.
But, 10-20 yield numbers such as 1, 14, 6, 12, 13. Here, the output is outside of the boundaries as well as in-between them. What am I doing wrong?
Thank you in advance for your help!

Answer 1

Using the c++11 random library.
You could use std::default_random_engine (or any other random engine) with std::uniform_int_distribution.
The values you pass to std::uniform_int_distribution will be the lower and upper bounds of your random range.
e.g.
#include <iostream>
#include <random>
int main()
{
const int nrolls = 100;
std::default_random_engine generator;
std::uniform_int_distribution<int> distribution(0,9);
int p[nrolls]={};
for (int i=0; i<nrolls; ++i)
{
p[i] = distribution(generator);
}
std::cout << "uniform_int_distribution (0,9):" << '\n';
for (int i=0; i<nrolls; ++i)
{
std::cout << i << ": " << p[i] << '\n';
}
}
If you wish to seed the random engine, create it like.
unsigned seed = std::chrono::system_clock::now().time_since_epoch().count();
std::default_random_engine generator(seed);

Answer 2

rand returns number between 0 and RAND_MAX. By taking modulo userEnd - userBeg + 1 the boundaries will be limited to 0 and userEnd - userBeg. If the random number should be within given boundaries, then userBeg should be added, so the calculus becomes
outPut = rand()%((userEnd - userBeg) + 1) + userBeg;

Answer 3

It's still experimental, but it is exactly what you want
http://en.cppreference.com/w/cpp/experimental/randint
#include <iostream>
#include <experimental/random>
int main()
{
int random_number = std::experimental::randint(100, 999);
std::cout << "random 3-digit number: " << random_number << '\n';
}
Edited: it's really random http://en.cppreference.com/w/cpp/numeric/random

Answer 4

For the rand to work you need to do:
#include <cstdlib>
int main() {
int Min = 103;
int Max = 113;
int Number = std::rand() % (Max + 1 - Min) + Min;
return 0;
}
I omitted the cin for clarity. (Max + 1 - Min) makes it inclusive. + Min sets the minimum value!
So stuffed into a function you may use:
int randint(int Min, int Max) {
return std::rand() % (Max + 1 - Min) + Min;
}
After testing it works with negative numbers as well as positive.
Sample outputs (with a loop and cout):
11-23
The number is: 13
The number is: 18
The number is: 14
The number is: 17
The number is: 18
The number is: 18
The number is: 23
The number is: 15
The number is: 11
The number is: 22
The number is: 22
The number is: 11
The number is: 22
The number is: 16
The number is: 14
The number is: 21
The number is: 16
The number is: 19
The number is: 15
The number is: 13
The number is: 17
1239-1242
The number is: 1240
The number is: 1242
The number is: 1239
The number is: 1241
The number is: 1241
The number is: 1241
The number is: 1239
The number is: 1240
The number is: 1240
The number is: 1240
The number is: 1242
The number is: 1240
The number is: 1242