c ++创建一个0.1到10之间的随机小数

Question

How would I do this?

This is my attempt of doing so:

srand (time(NULL));
seed = ((double)rand()) / ((double)RAND_MAX) * 10 + 0.5;

Also what is the way of creating a random integer between 0 and some int x. [0,x]

Answer 1

The C++11 way:

#include <random>

std::random_device rd;
std::default_random_engine generator(rd()); // rd() provides a random seed
std::uniform_real_distribution<double> distribution(0.1,10);

double number = distribution(generator);

If you only want integers, use this distribution instead:

std::uniform_int_distribution<int> distribution(0, x);

C++11 is really powerful and well-designed in this respect. The generators are separate from the choice of distribution, ranges are taken into account, thread safe, performance is good, and people spent a lot of time to make sure it's all correct. That last part is harder to get right than you think.

Answer 2

srand (time(NULL));
seed = ((double)rand()) / ((double)RAND_MAX) * 9.9 + 0.1;

To show up to 2 decimal places:

printf("%.2lf\n", seed);

If the x you need is smaller than RAND_MAX , then use

seed = rand() % (x+1);

to generate an integer in [0, x] .

Answer 3

#include <iostream>
#include <algorithm>
#include <vector>
#include <ctime>
#include <cstdlib>

using namespace std;

float r(int fanwei)
{
    srand( (unsigned)time(NULL) ); 
    int nTmp =  rand()%fanwei;
    return (float) nTmp / 10;
}

int main(int argc, const char * argv[])
{
    cout<<r(100)<<endl; 
    return 0;
}