What you get with: (0:n-1)*n + 1:n
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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
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Accepted Answer
Patrick Kalita
on 12 Apr 2011
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
8 Comments
Patrick Kalita
on 12 Apr 2011
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)."
More Answers (3)
Sean de Wolski
on 12 Apr 2011
And the transpose acts as expected:
>> ((0:n-1)*n)'
ans =
0
4
8
12
>> ans+(1:n)'
ans =
1
6
11
16
Sean de Wolski
on 12 Apr 2011
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
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