Random number c++ in some range [duplicate]

2019-01-10 17:03发布

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.

Thanks

标签: c++ random
6条回答
可以哭但决不认输i
2楼-- · 2019-01-10 17:24
int range = max - min + 1;
int num = rand() % range + min;
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混吃等死
3楼-- · 2019-01-10 17:29
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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倾城 Initia
4楼-- · 2019-01-10 17:34
float RandomFloat(float min, float max)
{
    float r = (float)rand() / (float)RAND_MAX;
    return min + r * (max - min);
}
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Viruses.
5楼-- · 2019-01-10 17:43

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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看我几分像从前
6楼-- · 2019-01-10 17:46

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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倾城 Initia
7楼-- · 2019-01-10 17:50

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 eng(rd()); // seed the generator
    std::uniform_int_distribution<> distr(25, 63); // define the range

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