NaN or Inf arrays preallocation

Question

Array indexing can be used for efficient array preallocation. For instance

2(ones(1, 3))
ans =

   2   2   2

but this does not work with NaN or Inf

NaN(ones(1, 3))
ans = NaN

Why ?

Answer 1

NaN and Inf look like special variables, when used without parenthesis.

But they are actually functions.

NaN(ones (1, 3)) expands to NaN ([1, 1, 1]) which apparently is evaluated like NaN (1, 1, 1). That is to a 1x1x1 array, which has only a single element.

The correct way to initialize a 1x3 NaN array is

NaN (1, 3)

Same for Inf.

---

Following @carandraug suggestion, here is a slight digression.

One might also use NaN ()(ones(1, 3)).

In this expression, NaN () evaluates to the NaN scalar value (not a function anymore). ones(1, 3) evaluates to [1, 1, 1].

So an intermediate step could be read as <NaN scalar value>([1 1 1]).

Then remember how indexing works. Indexing of an array A with an array of integers indexes is written A(indexes). For instance

A([i1, i2, i3])

This prepares an array of the same size as indexes (1x3 here). Each element of this new array will get the value of the element of A having the index given by the corresponding element of indexes. That is

[A(i1), A(i2), A(i3)]

So the result of 2(ones (1, 3)), i.e. 2([1, 1, 1]) is obviously [2(1), 2(1), 2(1)]. i.e. [2, 2, 2]. (Remembering that a scalar can be interpreted as a single element array. So 2(1) means first element of the array [2], which is 2).

Similarly, the intermediate step <NaN scalar value>([1 1 1]) is finally transformed in

[<NaN scalar value>, <NaN scalar value>, <NaN scalar value>]

or simply [NaN, NaN, NaN].