- Double has 2x more precision then float.
- float is a 32 bit IEEE 754 single precision Floating Point Number1 bit for the sign, (8 bits for the exponent, and 23* for the value), i.e. float has 7 decimal digits of precision.
- Double is a 64 bit IEEE 754 double precision Floating Point Number (1 bit for the sign, 11 bits for the exponent, and 52* bits for the value), i.e. double has 15 decimal digits of precision.
Let’s take a example(example taken from here) :
For a quadratic equation x2 – 4.0000000 x + 3.9999999 = 0, the exact roots to 10 significant digits are, r1 = 2.000316228 and r2 = 1.999683772
// C program to demonstrate
// double and float precision values
#include <stdio.h>
#include <math.h>
// utility function which calculate roots of
// quadratic equation using double values
void double_solve(double a, double b, double c)
{
double d = b*b - 4.0*a*c;
double sd = sqrt(d);
double r1 = (-b + sd) / (2.0*a);
double r2 = (-b - sd) / (2.0*a);
printf("%.5f\t%.5f\n", r1, r2);
}
// utility function which calculate roots of
// quadratic equation using float values
void float_solve(float a, float b, float c)
{
float d = b*b - 4.0f*a*c;
float sd = sqrtf(d);
float r1 = (-b + sd) / (2.0f*a);
float r2 = (-b - sd) / (2.0f*a);
printf("%.5f\t%.5f\n", r1, r2);
}
// driver program
int main()
{
float fa = 1.0f;
float fb = -4.0000000f;
float fc = 3.9999999f;
double da = 1.0;
double db = -4.0000000;
double dc = 3.9999999;
printf("roots of equation x2 - 4.0000000 x + 3.9999999 = 0 are : \n");
printf("for float values: \n");float_solve(fa, fb, fc);
printf("for double values: \n");
double_solve(da, db, dc);
return 0;
}
Output:
roots of equation x2 - 4.0000000 x + 3.9999999 = 0 are :
for float values:
2.00000 2.00000
for double values:
2.00032 1.99968
