# A problem with function handle

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Mohammad Shojaei Arani on 6 Nov 2022
Hello,
I have a simple problem but I do not understand why function handle behaves like this!!! I explain by a simple example in bellow:
>> x1=linspace(-pi,pi,20);y1=sin(x1);f=@(x)interp1(x1,y1,x);f(5)
ans =
NaN
which makes sense. But, when I type
>> x1=linspace(-pi,pi,20);y1=sin(x1);f=@(x)interp1(x1,y1,x).*(x>=-pi & x<=pi)+0.*(x<-pi | x>pi);f(5)
ans =
NaN
and this does not make sense to me.
Any idea?
Babak

John D'Errico on 6 Nov 2022
Edited: John D'Errico on 6 Nov 2022
NaNs are like wire coathangars, they multiply. Or perhaps evil zombies is a better description. They slime everything they touch, turning everything into new NaNs, propagating NaNs almost everywhere. Evil indeed. ;-)
For example, NaN plus ANYTHING = NaN.
NaN + 2
ans = NaN
NaN times ANYTHING = NaN. Even zero.
NaN*0
ans = NaN
So what did you do? You were hoping the trick of putting a test inline, that a multiply by zero would kill off the NaNs? Yeah, right. The zombies still survived.
x1=linspace(-pi,pi,20);
y1=sin(x1);
extrapval = 42;
f=@(x)interp1(x1,y1,x,[],extrapval);
f(5)
ans = 42
I cannot understand why interp1 uses NaN as the default for extrapolation, as 42 has been proven to be the answer to all questions. But if you really want zero to be returned, you can tell interp1 to do that too.
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Mohammad Shojaei Arani on 6 Nov 2022
Thanks John!
I came up by an idea to solve it. But, your way of solving the problem is much more elegant than mine!