We closed this forum 18 June 2010. It has served us well since 2005 as the ALPHA forum did before it from 2002 to 2005. New discussions are ongoing at the new URL http://forum.processing.org. You'll need to sign up and get a new user account. We're sorry about that inconvenience, but we think it's better in the long run. The content on this forum will remain online.
Index › Programming Questions & Help › Syntax Questions › float g = pow(-0.21027565, 1.2674413); => NaN
‹ Previous Topic | Next Topic ›
Page Index Toggle Pages: 1
float g = pow(-0.21027565, 1.2674413); => NaN (Read 286 times)
maurito
float g = pow(-0.21027565, 1.2674413); => NaN
Oct 22nd, 2008, 6:24am  
Hi everyone

I was programming and I got into a situation where I had to rise a float number to the power of another float number. It all was going fine until the base number became negative as in the example below:

float g = pow(-0.21027565, 1.2674413);
println(g);

This ended up printing a NaN message. I fixed my problem with an if statement and a -1 multiplication, yet I don't seem to understand where does the error come from.
If any of you know, it would be nice to hear it.

Cheers
Back to top
 
 
davbol
Re: float g = pow(-0.21027565, 1.2674413); => N
Reply #1 - Oct 22nd, 2008, 5:10pm  
pow(a,b) returns NaN for negative a and fractional b -- the result is complex and plain old pow() can't do it.  perhaps (?) it's clearer to see that pow(a,b) implies exp(b*log(a)), and log(a) is undefined for negative values. btw, also note that -pow(a,b) != pow(-a,b).  does that help?
Back to top
 
WWW  
maurito
Re: float g = pow(-0.21027565, 1.2674413); => N
Reply #2 - Oct 24th, 2008, 7:51am  
Thanks davbol that did, help.

I guess doing it right would imply using imaginary numbers, but since I want to use the results as weights to a function, using -pow(a,b) instead of pow(-a,b) should work fine.

Cheers
Back to top
 
 
Page Index Toggle Pages: 1
‹ Previous Topic | Next Topic ›