What you get with: (0:n-1)*n + 1:n

What you get with:

n = 4;
(0:n-1)*n + 1:n % 1 2 3 4

I was expecting:

(0:n-1)*n + (1:n) % 1 6 11 16

Answer 1

[Patrick Kalita]

Two things going on here.

1) Addition is done before the colon operator. Compare this:

 >> 1 + 1:4

 ans =

      2     3     4

and this:

 >> 1 + (1:4)

 ans =

      2     3     4     5

That means that (0:n-1)*n + 1 is evaluated to a vector and that vector becomes the first input to another colon operator.

2) When a vector is given to the colon operator, the first element is used:

 >> (1:4):3

 ans =

      1     2     3

[Oleg Komarov]

What about the behaviour that Sean outlined for column vector against row colon expansion with no brackets?

Is this behaviour documented?

[Matt Fig]

Sean de only points out the addition of two column vectors.

[Patrick Kalita]

@Oleg, I'm not sure what you're referring to. Sean's example shows one column vector being generated by the first statement. Then it is added to another column vector in the second statement. It is the same as your second statement but it uses column vectors instead of row vectors.

[Oleg Komarov]

@Matt, Patrick: right! I read column vectors but decided subconsciously to make it more complicated.

[Oleg Komarov]

Still can't find where this behavior is documented...

[Walter Roberson]

Oleg, with the various comments at various times, I am not certain which behaviour it is you are asked about the documentation for?

http://www.mathworks.com/help/techdoc/matlab_prog/f0-40063.html#f0-38155 shows that addition is done before colon.

[Patrick Kalita]

My second point is documented here: http://www.mathworks.com/help/techdoc/ref/colon.html

It says "If you specify nonscalar arrays, MATLAB interprets j:i:k as j(1):i(1):k(1)."

[Oleg Komarov]

@Walter: tried to look for 'operator order' and didn't think of 'precedence' (cost of being non-native). Thanks!

Answer 2

[Sean de Wolski]

And the transpose acts as expected:

>> ((0:n-1)*n)'

ans =

     0
     4
     8
    12

>> ans+(1:n)'

ans =

     1
     6
    11
    16

[Oleg Komarov]

Here the fact that 1:n is enclosed in parentheses evaluates the addition after the colon expansion.

Answer 3

[Sean de Wolski]

I'm able to replicate this. Mac OSX R2009b.

It also fails if I break it between lines:

>> ((0:n-1)*n)
ans =
   0     4     8    12
>> ans+1:n
ans =
   1     2     3     4

Answer 4

[VIGNESH]

u = 0:1/n:1