为了表示浮点数,我们使用float , double和long double 。
有什么不同 ?
double的精度是float的2倍。
float是32位IEEE 754单精度浮点数1位符号,(8位为指数,23 *为值),即float具有7位十进制精度。
double是64位IEEE 754双精度浮点数(符号1位,指数11位,值52 *位),即double的精度为15位小数。
让我们举个例子(例子取自这里):
对于二次方程x2 – 4.0000000 x + 3.9999999 = 0 ,精确到10个有效数字的根是r1 = 2.000316228和r2 = 1.999683772
// C program to demonstrate
// double and float precision values
#include
#include
// 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:
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
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