Random number c++ in some range

Possible Duplicate:
Generate Random numbers uniformly over entire range

I want to generate the random number in c++ with in some range let say i want to have number between 25 and 63.

How can I have that?

c++
random

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int range = max - min + 1;
int num = rand() % range + min;

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最佳答案

You can use the random functionality included within the additions to the standard library (TR1). Or you can use the same old technique that works in plain C:

25 + ( std::rand() % ( 63 - 25 + 1 ) )

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float RandomFloat(float min, float max)
{
float r = (float)rand() / (float)RAND_MAX;
return min + r * (max - min);
}

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Use the rand function:

http://www.cplusplus.com/reference/clibrary/cstdlib/rand/

Quote:

A typical way to generate pseudo-random numbers in a determined range using rand is to use the modulo of the returned value by the range span and add the initial value of the range:


( value % 100 ) is in the range 0 to 99
( value % 100 + 1 ) is in the range 1 to 100
( value % 30 + 1985 ) is in the range 1985 to 2014

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int random(int min, int max) //range : [min, max]
{
static bool first = true;
if (first)
{
srand( time(NULL) ); //seeding for the first time only!
first = false;
}
return min + rand() % (( max + 1 ) - min);
}

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Since nobody posted the modern C++ approach yet,

#include <iostream>
#include <random>
int main()
{
std::random_device rd; // obtain a random number from hardware
std::mt19937 gen(rd()); // seed the generator
std::uniform_int_distribution<> distr(25, 63); // define the range


for(int n=0; n<40; ++n)
std::cout << distr(gen) << ' '; // generate numbers
}