C / C ++中float和double的区别-IDC帮帮忙

为了表示浮点数,我们使用floatdoublelong double

有什么不同 ?

精度比浮动精度高2 

float是符号的32位IEEE 754单精度浮点数1位(指数为8位,值为23 *),即float具有7个十进制数字的精度。

double是64位IEEE 754双精度浮点数(符号为1位,指数为11位,值为52 *位),即double有15位十进制数字。

让我们举一个例子(例子来自这里):
对于二次方程x2 - 4.0000000 x + 3.9999999 = 0,精确根到10位有效数字是,r1 = 2.000316228和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;

输出:

方程x2  -  4.0000000 x + 3.9999999 = 0的根是: 
对于浮点值: 
2.00000 2.00000
对于双值: 
2.00032 1.99968