Typecasting and math function

**wozaa** · 03-18-2009

Hello

I am very confused at my problem. I have a piece of code that sometimes offers the correct result, and sometimes not.

The code is similar to this

Code:

#include "cmath"
#include "iostream"
using namespace std;

int log(int a){
    return (int)log2(a+1);
}

int main(){
int a =7;
cout << log(a) << endl;
}

for a = 7 this returns 2 instead of 3.
for a = 15 it returns the correct value of 4
it seems to alternate for other powers of two.

I programmed using g++ on ubuntu which gave me some form of converting double to int warning which I managed to suppress by typecasting (int). However when I went to test my program on windows using cygwin the (int) seems to be the cause of error.

Code:

in windows:
int a = 7;
cout << log2(a+1) << endl; gives 3 
cout << (int)log2(a+1) << endl; gives 2

What is the cause of this?

thanks

**vart** · 03-18-2009

2.99...99
printed as double will probalbly shown 3
(int)2.99...99 will give you 2

**EVOEx** · 03-18-2009

vart is right. The solution would be to do:

Code:

return (int)(log2(a+1) + epsilon);

Where epsilon is a small value, eg. 1e-12.

**wozaa** · 03-18-2009

Thanks you are right adding epsilon works.

Why would the math function return 2.999 instead of 3

Can I blame the compiler?

**vart** · 03-18-2009

Originally Posted by wozaa

Thanks you are right adding epsilon works.

Why would the math function return 2.999 instead of 3

Can I blame the compiler?

float arithmetic is not precise. to calculate log2 probably some long sum is used, which is stopped as soon as some predefined precision is achieved. so any float point operaton result could be a little less or bigger than presize result. you should take this into account.

This is one of the resons not to use == on float types

**wozaa** · 03-18-2009

Originally Posted by vart

float arithmetic is not precise. to calculate log2 probably some long sum is used, which is stopped as soon as some predefined precision is achieved. so any float point operaton result could be a little less or bigger than presize result. you should take this into account.

This is one of the resons not to use == on float types

Thanks, makes sense.

**matsp** · 03-18-2009

Originally Posted by vart

float arithmetic is not precise. to calculate log2 probably some long sum is used, which is stopped as soon as some predefined precision is achieved. so any float point operaton result could be a little less or bigger than presize result. you should take this into account.

This is one of the resons not to use == on float types

Actually, on a processor with floating point, it's likely that log2 is simply log(x) / log(2) - but either way, the result is not EXACT.

If you really just want log2 to the nearest integer value, you could use a method of figuring out the highest bit - there are several ways to do that, but this works:

Code:

int log(int x)
{
    int cnt = -1;  
    x ++;   // Make it x + 1 first. 
    do
    {
       cnt++;
       x >>= 1;
    } while(x);
    return cnt;
}

Note that the above code doesn't work for negative numbers (neither does log).
--
Mats

Thread: Typecasting and math function

Thread Tools

Search Thread

Display

Typecasting and math function