Trying to package anonymous function + function parameters in a structure
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Daniel Paulsen
on 23 Apr 2020
Commented: Daniel Paulsen
on 23 Apr 2020
I assumed this would work but it doesn't:
myStruct.A = 3;
myStruct.B = 7;
myStruct.myFunc = @(x) myStruct.A*x + myStruct.B;
test1 = myStruct.myFunc(1) %Returns 10
myStruct.B = 6;
test2 = myStruct.myFunc(1) %Still returns 10 but i want it to return 9
This is a trivial example but in my actual script the function i am working with it quite complicated; Imagine that this goes from Struct.A to Struct.Z and every one is a vector or matrix. This would get very tedious if I had to write the function with a @(x, A,B,C,..Z).
Looking for recomedations on how to acheive this.
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Accepted Answer
Steven Lord
on 23 Apr 2020
Your anonymous function takes a "snapshot" of the variable myStruct as it exists when the anonymous function is taken. The only way to update that "snapshot" is to recreate the anonymous function. You can see this using the debugging tool functions. As its documentation page states, don't try using this programmatically. It's intended to debugging and diagnosing problems.
myStruct.A = 3;
myStruct.B = 7;
myStruct.myFunc = @(x) myStruct.A*x + myStruct.B;
data = functions(myStruct.myFunc)
data.workspace{1}
data.workspace{1}.myStruct
Probably the easiest option, since you say you don't want to define an anonymous function with 27 inputs (and I wouldn't want to define such a function either) is to define one with two inputs.
myStruct.myFunc = @(x, myStruct) myStruct.A*x + myStruct.B;
myStruct.A = 3;
myStruct.B = 7;
myStruct.myFunc(1, myStruct) % 10
myStruct2 = myStruct;
myStruct2.B = 6;
myStruct.myFunc(1, myStruct2) % 9
This has an added benefit of allowing you to pass the same struct array into multiple functions. Each one uses the pieces of data it wants and ignores those pieces it doesn't.
myStruct2.C = 46;
myStruct2.myFunc2 = @(x, myStruct) myStruct.C-x./myStruct.A;
myStruct2.myFunc2(12, myStruct2) % ignores myStruct2.B entirely
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