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Difference between float and double in C/C++ - GeeksforGeeks

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Difference between float and double in C/C++

  • Difficulty Level : Easy
  • Last Updated : 26 Apr, 2018

For representing floating point numbers, we use floatdouble and long double.

What’s the difference ?

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    2.00000for double values:2.00032    1.99968

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