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

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