Substitute variables in function handle

Hello,
I have two implementations.
  1. Create functions based on the parameter
  2. Pass this functions to another function and evaluate them there
I'm creating functions like this:
for i=1:n/2-1
funk{i} = @(x)2*x(i)-2+400*x(i)^3-400*x(i)*x(i+1);
end
I want this to create multiple functions (depends on n) with i to be superset by a number, but WITHOUT being solved right away as i want to hand it over to another function in which this should get solved.
How can I achieve this?
Thanks in advance

Antworten (1)

Walter Roberson
Walter Roberson am 24 Jan. 2022

0 Stimmen

What you have already done should work.
Although all of the function handles will look the same, they would have captured different values of i
For example,
for i = 1 : 2
funk{i} = @(x)2*x(i)-2+400*x(i)^3-400*x(i)*x(i+1);
end
ws1 = functions(funk{1}).workspace{1}
ws1 = struct with fields:
i: 1
ws2 = functions(funk{2}).workspace{1}
ws2 = struct with fields:
i: 2
Notice how the information embedded within the function handles funk{1} and funk{2} is different -- they have different embedded i values.

2 Kommentare

So I can simply pass this to another function like:
for i = 1 : 2
funk{i} = @(x)2*x(i)-2+400*x(i)^3-400*x(i)*x(i+1);
end
function test(fun)
test(funk)
This returns an error that Array indices must be positive integers or logical values in my implementation of the calling function.
Torsten
Torsten am 24 Jan. 2022
Bearbeitet: Torsten am 24 Jan. 2022
x = [1 2 3 4 5 6];
for i = 1 : 2
funk{i} = @(x)2*x(i)-2+400*x(i)^3-400*x(i)*x(i+1);
end
test(funk,x)
function test(fun,x)
fun{1}(x)
fun{2}(x)
end

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R2021b

Gefragt:

am 24 Jan. 2022

Bearbeitet:

am 24 Jan. 2022

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