Thread: How to generate a random number in the interval of -1 and 1?

Originally Posted by flp1969

That's interesting, Malcom! I completely forgot about linear interpolation!

Code:

#include <stdlib.h>
#include <float.h>

#define lerp(min_, max_, norm) \
  ( (min_)*(1.0 - (norm)) + (max_)*(norm) )

double symmetrical_rand ( void )
{
  double d;

  d = ( double ) rand() / RAND_MAX;
  return lerp ( -1.0 + DBL_EPSILON, 1.0 - DBL_EPSILON, d );
}

There is one more number between 1.0-DBL_EPSILON and 1.0:

Code:

#include <float.h>
#include <stdio.h>


void printhex(void *x, size_t s) {
  for(size_t i = 0; i < s; i++)
    printf("%02X ",((unsigned char *)x)[i]);
  putchar('\n');
}


int main(int argc, char *argv[]) {
   double a = 1.0;
   double b = 1.0-DBL_EPSILON/2;
   double c = 1.0-DBL_EPSILON;
   printf("%e\n", DBL_EPSILON);
   printf("%32.30e\n", a);
   printf("%32.30e\n", b);
   printf("%32.30e\n", c);
   printhex(&a, sizeof(a));
   printhex(&b, sizeof(b));
   printhex(&c, sizeof(c));


}

Gives the following output:

Code:

2.220446e-16
1.000000000000000000000000000000e+00
9.999999999999998889776975374843e-01
9.999999999999997779553950749687e-01
00 00 00 00 00 00 F0 3F
FF FF FF FF FF FF EF 3F
FE FF FF FF FF FF EF 3F

Guys, I am sorry to be late to deliver THANKS to everyone. Appreciated so much that I have learned from your code. Thanks again.

Originally Posted by hamster_nz

There is one more number between 1.0-DBL_EPSILON and 1.0

Ahhh... of course, since we change the scale from 2⁰ to 2⁻¹! Nice catch.