index out of bounds error

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soloby on 1 Jul 2015
Answered: Brendan Hamm on 1 Jul 2015
Hey all, my code looks something like this.
x = -10:0.1:10;
f1 = trapmf(x,[-2 0 0 2]);
f2 = trapmf(x,[1 3 3 5]);
kvals = 0.05:0.05:1;
for kidx = 1 : length(kvals)
k = kvals(kidx);
index1 = find(abs(f1-k)<10^(-10));
Xidx1{kidx} = x(index1);
Xidx1{20} = [0 0];
index2 = find(abs(f2-k)<10^(-10));
Xidx2{kidx} = x(index2);
Xidx2{20} = [0 0];
A(kidx) = Xidx1{kidx}(1);
B(kidx) = Xidx1{kidx}(2);
C(kidx) = Xidx2{kidx}(1);
D(kidx) = Xidx2{kidx}(2);
It works perfectly for what it is, getting lower and upper bounds of my functions f1 and f2.
But for some reason if I change my f1 or f2 functions to something else, I am getting a number of elements error?
Like f2 = trapmf(x,[1 3 3 4])
will give me an error, and I can't seem to figure out why that is?
the exact error message is
Attempted to access Xidx2.%cell(2); index out of bounds because
Error in (line 22)
D(kidx) = Xidx2{kidx}(2);

Accepted Answer

Brendan Hamm
Brendan Hamm on 1 Jul 2015
You are using the find function and then assigning values based on that to the variable Xidx2{kidx}. If you use the debugger with the change you mentioned you will notice that on the first run there is exactly one return value for the line:
index2 = find(abs(f2-k)<10^(-10));
Therefore the value being passed in the following assignment is a scalar:
Xidx2{kidx} = x(index2);
Therefore you get an error in the indexing that follows:
D(kidx) = Xidx2{kidx}(2);
There is nothing in this code which will guarantee that there are multiple values which satisfy the inequality:
You could use an if statement or try-catch block to perform this check for you.
When you encounter these errors, use the debugger , it is your friend.

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