Well, first thing first.
1. What does this warning mean?
This code is correct in terms of syntax and it will execute correctly returning the expected result: the ii-th element of x will contain the value foo( ii ).
However, before this small piece of code runs, the variable x is not defined. Now, when the loop starts, x(1) is assigned the value foo( 1 ), and so Matlab creates x as a length-1 array. At the second iteration x(2) is assigned the value foo( 2 ) and so Matlab needs to change x to be of length 2, and so on: x changes its length/size at each iteration.
2. Why is changing variable size every iteration is a bad thing?
Consider what happens in the background (in terms of memory allocation) when x changes its size every iteration: At each iteration Matlab needs to find a free memory space to host the new size of x. If you are lucky, there is enough free space right after x so all that happens is a change to the amount of memory allocated to x and writing the new value at the right spot.
However, if there is not enough free space just after x, Matlab has to find a new spot for all the ii-1 elements already in x, allocate this new space for x, copy all ii-1 values already in x to the new spot, and free the old spot x used. This allocate-copy-free operations happening in the background can be extremely time consuming, especially when x is large.
3. How can this problem be solved?
The simplest solution is to pre-allocate all the space x needs before the loop:
x = zeros(1,n);
for ii=1:n
x( ii ) = foo( ii );
end
By pre-allocating we ascertain that x is allocated all the memory it requires up-front, thus no costly memory allocation/copy is needed when the loop is executing.
An alternative cool solution to the problem
If you are too lazy (like me) and don’t want to pre-allocate you can simply:
for ii=n:-1:1
x( ii ) = foo( ii );
end
This way, the first time x is assigned a value it is assigned to its n-th element (the last one) and therefore Matlab immediately allocates room for all n elements of x.
Cool!