How must I declare a function handle that refers to a function you've written? What syntax is necessary?

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Daniel
Daniel 2012-2-7
I keep getting the error, "??? Output argument [...] not assigned during call to" my function psi_en whenever I try any of the following commands:
Case 1:
quad_fun = @(E) mass_ab_E(E).*exp(-depth*mass_E(E)*rho_water) ...
.*psi_en(E,spectrum);
max_energy = spectrum(length(spectrum),1);
y = quad(quad_fun,0,max_energy);
Case 2:
quad_fun = @(E) mass_ab_E(E).*exp(-depth*mass_E(E)*rho_water);
max_energy = spectrum(length(spectrum),1);
y = quad(@(E) quad_fun.*psi_en(E,spectrum),0,max_energy);
Case 3:
psi_E = @(E) psi_en(E,spectrum); % error points to both lines as problematic
quad_fun = @(E) mass_ab_E(E).*exp(-depth*mass_E(E)*rho_water).*psi_E(E);
I think it is angry that I'm trying to define a function handle that refers to a function I've written! I can execute the function directly, e.g. "psi_en(0.02,spectrum)" returns "ans = 0.0041", but it takes issue with this function handle declaration. What must I do, and why is it unhappy?!
psi_en.m contains the following; it is intended to assign values to psi_en based on where the quad integration variable E falls within the spectrum histogram (first column is the energy bin, second column is the energy's weighting factor psi_en):
function y = psi_en(en,spect)
if en < 0
error('energy must be positive!')
elseif en == 0
y = 0;
else
if (en > 0) & (en <= spect(1,1))
y = spect(1,2);
end
for i = 1:length(spect)-1
if (en > spect(i,1)) & (en <= spect(i+1,1))
y = spect(i+1,2);
end
end
if en > spect(length(spect))
error('Energy surpasses data!')
end
end

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采纳的回答

Walter Roberson
Walter Roberson 2012-2-7
You have flows of control where y is not set, or at least not obviously so.
When you have a function that returns a value, you should make sure that the output value is always defined (except possibly in cases you are going to error() out of.) You could initialize y to inf for example.
  2 个评论
Daniel
Daniel 2012-2-7
My problem seems more than that: I initialized y = 0 before entering the if statements, and now it's giving only 0. It appears not to be using the if statements; that is, it seems I am not properly calling the function. "psi_E = @(E) psi_en(E,spectrum);", "quad_fun = @(E) mass_ab_E(E).*exp(-depth*mass_E(E)*rho_water).*psi_E(E);" and "y = quad(quad_fun,0,max_energy);" -- it is not working as I think it should. Why is it so difficult to use this quad function?!
Walter Roberson
Walter Roberson 2012-2-7
quad passes a vector argument -- your E will be a vector, and that will reach your function psi_en through the argument en. You then do tests suitable for _scalars_ on the _vector_ . When "if" is applied to vector (or array) of values, the test is considered true if and only if _every_ value in the vector (or array) is non-zero -- if *all* of the values meet the condition.
If you are only getting the 0 output, the implication is that none of your "if" statements was true for all elements in the vector.
You need to either switch over to using a loop over the "en" values, or else switch over to using logical indexing (a better approach.)

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更多回答（1 个）

Kevin Holst
Kevin Holst 2012-2-7
Is 'spectrum' actually getting passed into your function psi_en? It looks like, if it isn't and 'en' in passed in correctly, 'y' would never be set.
  3 个评论
显示 1更早的评论
Kevin Holst
Kevin Holst 2012-2-7
I don't think the problem lies in assigning a function handle to to an expression that includes the function, there's something else at play. Which line of code in your main function is producing the error?
Daniel
Daniel 2012-2-7
It's the lines that involve the function handle. Assigning a value at the beginning to later be overwritten within the if statements solves the problem, as Roberson suggested, but it appears still not to be "working" as I like -- that is, psi_en is then always that value, apparently...

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