# Generating random number between [-1, 1] in C?

Use -1+2*((float)rand())/RAND_MAX

rand() generates integers in the range [0,RAND_MAX] inclusive therefore, ((float)rand())/RAND_MAX returns a floating-point number in [0,1]. We get random numbers from [-1,1] by adding it to -1.

EDIT: (adding relevant portions of the comment section)

On the limitations of this method:

((float)rand())/RAND_MAX returns a percentage (a fraction from 0 to 1). So since the range between -1 to 1 is 2 integers, I multiply that fraction by 2 and then add it to the minimum number you want, -1. This also tells you about the quality of your random numbers since you will only have RAND_MAX unique random numbers.

If all you have is the Standard C library, then other people's answers are sensible. If you have POSIX functionality available to you, consider using the drand48() family of functions. In particular:

#define _XOPEN_SOURCE 600  /* Request non-standard functions */
#include <stdlib.h>

double f = +1.0 - 2.0 * drand48();
double g = -1.0 + 2.0 * drand48();


Note that the manual says:

The drand48() and erand48() functions shall return non-negative, double-precision, floating-point values, uniformly distributed over the interval [0.0,1.0).

If you strictly need [-1.0,+1.0] (as opposed to [-1.0,+1.0)), then you face a very delicate problem with how to extend the range.

The drand48() functions give you considerably more randomness than the typical implementation of rand(). However, if you need cryptographic randomness, none of these are appropriate; you need to look for 'cryptographically strong PRNG' (PRNG = pseudo-random number generator).

For starters, you'll need the C library function rand(). This is in the stdlib.h header file, so you should put:

#include <stdlib.h>


near the beginning of your code. rand() will generate a random integer between zero and RAND_MAX so dividing it by RAND_MAX / 2 will give you a number between zero and 2 inclusive. Subtract one, and you're onto your target range of -1 to 1.

However, if you simply do int n = rand() / (RAND_MAX / 2) you will find you don't get the answer which you expect. This is because both rand() and RAND_MAX / 2 are integers, so integer arithmetic is used. To stop this from happening, some people use a float cast, but I would recommend avoiding casts by multiplying by 1.0.

You should also seed your random number generator using the srand() function. In order to get a different result each time, people often seed the generator based on the clock time, by doing srand(time(0)).

So, overall we have:

#include <stdlib.h>
srand(time(0);
double r = 1.0 * rand() / (RAND_MAX / 2) - 1;


I had a similar question a while back and thought that it might be more efficient to just generate the fractional part directly. I did some searching and came across an interesting fast floating point rand that doesn't use floating point division or multiplication or a int->float cast can be done with some intimate knowledge of the internal representation of a float:

float sfrand( void )
{
unsigned int a=(rand()<<16)|rand();  //we use the bottom 23 bits of the int, so one
//16 bit rand() won't cut it.
a=(a&0x007fffff) | 0x40000000;

return( *((float*)&a) - 3.0f );
}


The first part generates a random float from [2^1,2^2), subtract 3 and you have [-1, 1). This of course may be too intimate for some applications/developers but it was just what I was looking for. This mechanism works well for any range that is a power of 2 wide.