difference in brackets (, [ or no brackets?
显示 更早的评论
I tried typing in a simple array
x= 0:2:10
x=[0:2:10]
x=(0:2:10)
they all give me the same answer whats the difference then?
采纳的回答
James Tursa
2016-10-18
编辑：James Tursa
2016-10-18
4 个投票
0:2:10 produces a row vector of the numbers 0,2,4,6,8,10.
In the [0:2:10] and (0:2:10) versions, the [ ] and ( ) are simply redundant and unnecessary, since what is inside the [ ] and ( ) is already a row vector.
In general, if you have only one argument to the [ ] or ( ) like above, the [ ] or ( ) are unnecessary. If you have more than one argument, then the [ ] actually does something. E.g.
x = [0,2,4,6,8,10]
In this case there are actually 6 separate arguments to the [ ], so the [ ] does something in this case (concatenates the 6 inputs into a single row vector).
1 个评论
(0:2:10) prioritizes the calculation of 0:2:10 . For example,
1 + 0:2:10
1 + (0:2:10)
The + operator normally has higher priority than the : operator, so 1 + 0:2:10 is interpreted as (1+0):2:10 which is 1:2:10
[0:2:10] builds the vector 0 2 4 6 8 10 in memory, but then calls the horzcat() operation, as if you had used horcat(0:2:10) . horzcat() with just a single input just returns the same input, so it is valid but uses time that is not productive.
更多回答（2 个）
Image Analyst
2016-10-18
0 个投票
No difference - they're ignored. Only difference would be if you used braces {}, which would turn the result into a "cell". See the FAQ for info on that: http://matlab.wikia.com/wiki/FAQ#What_is_a_cell_array.3F
ogaba-oche david
2023-5-10
0 个投票
(a+A)/2 keeps telling error, when a = 8 and A = -2
1 个评论
Not for me.
a = 8;
A = -2;
(a+A)/2
Please show us the full and exact text (everything in red) of the error message you received when you performed that operation. You may also want to show us more information about the variables a and A.
whos a A
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