use of parentheses confusion

how does the code operate differently between first and 2nd line?

10+1:5*3

ans = 1×5

11 12 13 14 15

10+(1:5*3)

ans = 1×15

11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Answer 1

[Kevin Holly]

In the first line, 10 is added to the 1

so you end up with

11:15 %This creates an array starting from 11 and ending at 15 at increments of 1 (1 is default if not specified

ans = 1×5

11 12 13 14 15

This is the same as

11:1:15 % start:increment:end

ans = 1×5

11 12 13 14 15

The second one creates creates an array from 1 to 15 and then adds 10

1:15 % same as 1:1:15

ans = 1×15

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

(1:15)+10

ans = 1×15

11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Answer 2

[John D'Errico]

This is a question about operator precedence, and we can find the rules here:

https://www.mathworks.com/help/matlab/matlab_prog/operator-precedence.html

In there, we see that colon operator is quite low on the scale. So it gets evaluated LAST. That means we can view this expression

10+1:5*3

ans = 1×5

11 12 13 14 15

as the same as

(10+1):(5*3)

ans = 1×5

11 12 13 14 15

Whew! I got lucky, and I was correct. MATLAB adds 10+1=11, as the lower limit for colon. The upper limit is 5*3=15.

In the second case, we know that : has the lowest order on that scale. So it will evaluate 5*3 as the upper limit for colon. But how about that 10? The 10 falls outside the parens. And parens is at the very top of the list. So what is inside the parens is evaluated before we add 10 to that result. So this expression generates the list 1:15. Then it adds 10 to every element in the list. So I expect to see the list 10:25 here.

10+(1:5*3)

ans = 1×15

11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Again, it works.