Matrix non-exceeding condition

Hello! I have a 1x15 matrix, the code is presented

[xdata,x]=min(Data); % Data 45x350   x=[ 16 16 16 31 16 0 16....]
a=[15 15 15 15 15....] % 1x350
 for i = 1:length(x)-1
    if x(i)+a(i)>a(i)
    ax=Data((x(i)-a(i):x(i)),i) ;
    ay=Data((x(i):x(i)+a(i),i)  ; 
    A{i}=ax;
    B{i}=ay;
    else x(i)+a(i)<a(i) ;
    ax=0 ;
    ay=0 ;   
    A{i}=ax;
    B{i}=ay;

    end
 end
%%%  x(i)+a(i) 31+15=46 

It so happened that in one value x (i) + a (i) exceeds the size of the matrix, can anyone tell me the condition how can I get around this?

[Adam]

Depends entirely what you want to happen in such a case. In theory it could happen in every position of your vector since you are adding 15 to a value that could be the biggest index in the vector.

You can clip x(i) + a(i) to the numel of x if you wish or you could ignore values outside the range, but we don't know what you want to happen where this occurs.

[Guillaume]

You will have to explain what your code is meant to do as it doesn't make much sense.

For example:

if x(i)+a(i)>a(i)

is the same as the much simpler:

if x(i) > 0

and since x are matrix indices (returned by min), you know in advance that it's always going to be true. So, what was the point of that?

Also,

else x(i)+a(i)<a(i) ;

is equivalent to:

else
    somethingdiscarded = (x(i) + a(i)) < a(i));

which is the same as

else
    somethingdiscarded = false;

Again, I'm sure that wasn't the intent.

[Lev Mihailov]

if x(i)+a(i)>size(Data)  % in my case it's 45
    ax=0;                % it was just written that it was zero
else x(i)+a(i)<a(i) 
    ax=0;

Something is needed

[Guillaume]

Something is needed

Yes, some much better explanation of what you're trying to do.

Once again,

else x(i)+a(i)<a(i) 
    ax=0;

is the same as

else 
    x(i)+a(i)<a(i)  %useless statement that will print 0 in the command window
    ax=0;

Perhaps, you meant something entirely different:

elseif x(i)+a(i)<a(i)
    ax=0;

but until you explain in words what you're trying to do, we can't know for sure.