"Array indices must be positive integers or logical values"

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for i=1:n
disp(i)
ti(i+1)=ti(i)+dt;
x1i(i+1)=x1i(i)+dt(V1i(i))
x2i(i+1)=x2i(i)+dt(V2i(i))
Using, this code I get the error message shown in the subject line, even if i change the updating to x1i(i)= ... this doesn't fix the problem.
Any advice?
  3 Comments
Rory Thiel
Rory Thiel on 22 Mar 2023
function [x1i,x2i, ti] = BasicEuler (RHS1,RHS2,initial, t0, tf, dt)
%create arrays for all variables which change over time
ti=zeros(1,1000);
x1i=zeros(1,1000);
x2i=zeros(1,1000);
V1i=zeros(1,1000);
V2i=zeros(1,1000);
ti(1)=t0;
%%define initial conditions (at time = 0s)
x1i(1)=initial(1);
x2i(1)=initial(2);
V1i(1)=initial(3);
V2i(1)=initial(4);
n=(tf-t0)/dt; %number of loops
i=0;
while max(ti)<tf
i=i+1;
ti(i+1)=ti(i)+dt;
x1i(i+1)=x1i(i)+dt(V1i(i));
x2i(i+1)=x2i(i)+dt(V2i(i));
%Continue on to calculate new V1 and V2 using RHS eqs
end
Here I have tried using a while loop instead but this doesn't resolve the issue
Which I am calling with
[x1i, x2i,ti] = BasicEuler (RHS1,RHS2,initial, t0, tf, dt);
after with all the required inputs in the main script
Rory Thiel
Rory Thiel on 22 Mar 2023
x1i(i+1)=x1i(i)+dt*(V1i(i));
Wasn't really a problem with indexing, I missed the * symbol after dt so it thought I was trying to search a dt array which is just a constant variable

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Answers (1)

Cris LaPierre
Cris LaPierre on 22 Mar 2023
Edited: Cris LaPierre on 22 Mar 2023
Your indices must be positive integer values (or logicals), as the error message states. It is therefore most likely either V1i(i) or V2i(i) are not returing an integer value, causing the error about invalid array indices when used to index into your variable dt.
a=1:3;
% This works
a(1)
ans = 1
% your error
a(1.5)
Array indices must be positive integers or logical values.

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