| 0,0 → 1,42 |
|---|
| #include <stdio.h> |
| #include <math.h> |
| |
| main() |
| { |
| int i; |
| for (i = 0; i < 10; i++) |
| { |
| printf("------------------------------------------------------\n"); |
| printf ( "remainder of 5.3 / 2 is %f\n", remainder (5.3,2) ); |
| printf ( "remainder of 18.5 / 4.2 is %f\n", remainder (18.5,4.2) ); |
| //remainder of 5.3 / 2 is -0.700000 |
| //remainder of 18.5 / 4.2 is 1.700000 |
| |
| printf ( "fmod of 5.3 / 2 is %f\n", fmod (5.3,2) ); |
| printf ( "fmod of 18.5 / 4.2 is %f\n", fmod (18.5,4.2) ); |
| // fmod of 5.3 / 2 is 1.300000 |
| // fmod of 18.5 / 4.2 is 1.700000 |
| |
| double param, fractpart, intpart, result; |
| int n; |
| |
| param = 3.14159265; |
| fractpart = modf (param , &intpart); |
| printf ("%f = %f + %f \n", param, intpart, fractpart); |
| //3.141593 = 3.000000 + 0.141593 |
| |
| param = 0.95; |
| n = 4; |
| result = ldexp (param , n); |
| printf ("%f * 2^%d = %f\n", param, n, result); |
| //0.950000 * 2^4 = 15.200000 |
| |
| param = 8.0; |
| result = frexp (param , &n); |
| printf ("%f = %f * 2^%d\n", param, result, n); |
| //8.000000 = 0.500000 * 2^4 |
| param = 50; |
| result = frexp (param , &n); |
| printf ("%f = %f * 2^%d\n", param, result, n); |
| } |
| } |