Taking in a function as an argument

I'm trying to make a function which takes in a function and then uses it. Right now my code looks like this:

function out = func(inputfunc, constant1, constant2)
inputfuncval1 = inputfunc(constant1);
...code...
end

When I then try to call it with a function like this:

out = func(@(x) testfunc(x),x1 ,x2);

It interpretets my input as an array instead of a function.

Anyone got any ideas on how to solve this issue?

[Guillaume]

Can you give us the full text of the error you see. The code you show should work although

out = func(@testfunc, x1, x2);

would be simpler.

[Jakob Grunditz]

To understand the error I must give you the whole code so... you're welcome

function svar = bisekt(f, x1, x2)
fx1 = f(x1);
fx2 = f(x2);

if fx1>0 && fx2<0
    
    if abs(x1-x2)>10^-6
        m = (x1+x2)/2;
        fm = f(m);
        if fm == 0
            svar = m;
            return 
        elseif fm>0
            bisekt(m, x2)
        elseif fm<0
            bisekt(x1, m)
        end
    else
        svar = x1;
        return
    end
        
        
elseif fx1<0 && fx2>0
    if abs(x1-x2)>10^-6
        m = (x1+x2)/2;
        fm = f(m);
        if fm == 0
            svar = m;
            return 
        elseif fm<0
            bisekt(m, x2)
        elseif fm>0
            bisekt(x1, m)
        end
    else
        svar = x1;
        return
    end
end
    
end

svar = bisekt(@(x) exp(x)+x^2-1, -2, -0.5)

error code
Array indices must be positive integers or logical values.

Error in bisekt (line 2)
fx1 = f(x1);

Error in bisekt (line 14)
            bisekt(m, x2)

Answer 1

[Guillaume]

Seems clear enough:

Error in bisekt (line 14)
            bisekt(m, x2)

Within bisekt, you call the function again, this time with input m which is a scalar value, not a function, and only one argument. So, the input f is then your m scalar, and of course, m(x2) is going to fail if x2 is anything but 1.

Perhaps, the call should be:

            bisekt(f, m, x2);

same for the other recursions...

[Jakob Grunditz]

Yes that seemed to be the problem x)... Thank you for spotting that.

Answer 2

[M]

How is testfunc defined ?

Here is a simple working example:

function out = func(inputfunc, constant1, constant2)
out = inputfunc(constant1, constant2);
end
out = func(@(x,y) x+y,1,2)
out =
     3

[Jakob Grunditz]

In my case I tried inputting the function:

inputfunc = @(x) exp(x)+x^2-1

that I then try to evaluate in two places constant1 and constant2 for later use in the function.

function out = func(inputfunc, constant1, constant2)
inputfuncval1 = inputfunc(constant1);
inputfuncval2 = inputfunc(constant2);
...code...
end
out = func(@(x) exp(x)+x^2-1, x1, x2)

[Guillaume]

As I commented in your question, if the above doesn't work, give us the full text of the error message.

With the following function:

function out = func(inputfunc, constant1, constant2)
out(1) = inputfunc(constant1);
out(2) = inputfunc(constant2);
end

I get:

>> out = func(@(x) exp(x)+x^2-1, 1, 2)
out =
          2.71828182845905          10.3890560989307

No problem there.