Thread: How to get the decimal part of a number...

Code:

#include <stdio.h>
#include <math.h>

int main(void)
{
    double integer, value = 237.3455, fraction = modf(value, &integer);
    printf("value = %g, integer = %g, fraction = %g\n",
           value, integer, fraction);
    return 0;
}

/* my output
value = 237.345, integer = 237, fraction = 0.3455
*/

... and if the number is on the trillions you will have a lot of ifing to do there. I don't think so...

modf() is the perfect solution whereas Sinkula's is the most pretty. I would only remove the C style casting in favor of C++ way.

try this:

Code:

float var1 = 123.456;

int tmp = (int)var1;

int var2 = var1 - tmp;

if ( var2 < 0 )
var2+1;     // cause the decimal part could be like 701 and 
                  // it will change the  tmp to 124 instead of 123

PS:
see this is what a real hacker does.... figures out a way of solving a problem that the others never thought off. dont misunderstand with cracker.

// cause the decimal part could be like 701 and

... and an hacker double checks his logic. A cast to int doesn't round. It truncates. Your if statement is unecessary.

Originally Posted by wyvern

see this is what a real hacker does.... figures out a way of solving a problem that the others never thought off.

Try posting something correct that solves the problem before bragging about it.

By the way, just the final nail on the coffin. I'm sure you don't mind, Wyvern. After all an hacker loves to learn new things too... It always produces 0 as the result.

Of course it does. After all, it's just Tonto's method, and then converting the decimal part gained to an int, thus guaranteeing 0 as the result. (Oh, and "var2+1" does nothing on its own. You need to assign the result.)