Hello, I wanted an inplace function to write a result into a subarray as follows:
X = zeros(10,5) fill!(X[:,3], 0.1) The X[:,3] is however not updated. X[:,3] = fill(0.1, 10) does the update as expected. Is this a desired behaviour ? --- Alternatively, I can do r = view(X,:,3) fill!(r, 0.1) This results in an updated column of X I wonder which is likely to be more efficient if used in a loop: X[:,3] = fill(0.1, 10) or r = view(X,:,3) fill!(r, 0.1) Thanks, Jan