Why is my function not working?

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Theobold
Theobold 2023년 2월 24일
댓글: Dyuman Joshi 2023년 2월 24일
I am very new to MATLAB, as in, started two days ago because I'm trying to translate some code from mathematica.
For context x0 = 1 and x,x2 are undefined in this function
function [f355] = f355(x,x0,x2)
%UNTITLED4 Summary of this function goes here
% Detailed explanation goes here
if x >= 0
f355 = (Omega00)*sqrt((1)+((Lambda355*x)/(pi*(Omega00))^2));
elseif (x<=(x2+x1+x0))&&(x>=x0)
f355 = NewBWaist(Omega00,x0,f,Lambda355)*sqrt((1)+(Lambda355*(x-x0-x1)/(pi(Omega01)^2)));
else x >= (x0+x1+x2)
f355 = NewBWaist(NewBWaist(Omega00,x0,f1U,Lambda355),x2,f2U,Lambda355)*sqrt((Lambda355(x-sumxU))/(pi*(Omega02)^2));
end
When I run "f355(10,1,20)" I get an error
"Index in position 1 exceeds array bounds. Index must not exceed 1.
Error in m118codeInprogress (line 111)
f355(10,1,20)"
Any advice would be most welcome. Thanks.
  댓글 수: 3
Theobold
Theobold 2023년 2월 24일
Apologies,
Omega00= 3.5
Lambda355= 0.000355
The rest of the variables are themselves defined by other functions with yet more variables. The other functions are all working though.
Dyuman Joshi
Dyuman Joshi 2023년 2월 24일
Rename your output variable name and run your code. As other functions are working properly, your code should work after the correction as well.

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Jan
Jan 2023년 2월 24일
이동: Jan 2023년 2월 24일
Avoid using the name of the function as name of the output. This is confusing only.
Which is the failing line? In this line f355 is used as a variable, not as a function.
See this example:
sin(0.1) % Fine
sin = 15; % Now sin is a variable
sin(0.1) % Error: 0.1 is no valid index
  댓글 수: 1
Theobold
Theobold 2023년 2월 24일
Right I see, I ran a calculation earlier (since deleted) with f355= so I guess it's now assigned as a variable?
Thanks for the advice.

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