How can I nest a function in a function?
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function AD = AD(n)
A = zeros(n);
for i=1:n
for j=1:n
A(i,j) = 1/(i+j-1);
end
end
Ainv = A\eye(n);
Eye = eye(n);
AD = Eye-Ainv*A;
end
function A = hilbert(n)
A = zeros(n);
for i=1:n
for j=1:n
A(i,j) = 1/(i+j-1);
end
end
end
I want to nest hilbert function in AD function, since hilbert is actually in AD. But I am getting "Unrecognized function or variable 'A'" error. What can I do here?
0 个评论
采纳的回答
Fangjun Jiang
2022-10-13
MyAD=AD(3)
function AD = AD(n)
A = hilbert(n);
Ainv = A\eye(n);
Eye = eye(n);
AD = Eye-Ainv*A;
end
function A = hilbert(n)
A = zeros(n);
for i=1:n
for j=1:n
A(i,j) = 1/(i+j-1);
end
end
end
2 个评论
Fangjun Jiang
2022-10-13
To be clear, hilbert(n) here is a 'local function'. There is no need to make it a 'nested function' based on its content.
See documents for their differences.
I think 'local function' is more suitable in this case.
更多回答(1 个)
John D'Errico
2022-10-13
编辑:John D'Errico
2022-10-13
See that I wrote hilbert differently. Feel free to use doubly nested loops there. But why?
As well, NEVER name a function the same thing as a variable in that function!!!!!!!!! NEVER. NEVER. Having said that three times, it must be true. You named the function AD, then returned a variable named AD. A BAD idea.
ComputeAd(5)
It works.
function AD = ComputeAd(n)
A = hilbert(n);
Ainv = A\eye(n);
Eye = eye(n);
AD = Eye-Ainv*A;
function A = hilbert(n)
[i,j] = meshgrid(1:n);
A = 1./(i+j-1);
end
end
3 个评论
John D'Errico
2022-10-13
Yes. You tried to call ComputeAD, but I named the function ComputeAd. Sorry about my choice of names.
Case sensitivity triumphs there. MATLAB told you it could not find that function, the clue you needed.
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