Using end as argument to function

Question

Michael 2018-3-23

2
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/390253-using-end-as-argument-to-function

评论： Michael 2018-3-23

采纳的回答： James Tursa

So today i noticed something i think is weird.

Say i define an anonyomous function.

f = @(x) ... Now if i input f(end) it will evaluate the function at f(1).

Is this a bug or is there a reason for this behaviour?

Regards Michael

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Answer 1

James Tursa 2018-3-23

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/390253-using-end-as-argument-to-function#answer_311590

编辑：James Tursa 2018-3-23

在 MATLAB Online 中打开

Weird, but it does match the doc for end:

"... end = (size(x,k)) when used as part of the kth index ..."

E.g.,

>> f = @(x)x+5
f = 
    @(x)x+5
>> size(f,1)
ans =
     1
>> f(1)
ans =
     6
>> f(end)
ans =
     6

f(end) appears as an indexing expression to MATLAB. Since the size of the f variable is 1x1, it uses 1 for end.

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James Tursa 2018-3-23

编辑：James Tursa 2018-3-23

在 MATLAB Online 中打开

Seems the parser processes the 'end' first and effectively does a replacement, without regard to whether the variable in question is a function handle or not. It does have a weird effect, however, in that the index itself gets magically turned into an argument for the function handle. E.g., using the same example:

>> f = @(x)x+5
f = 
    @(x)x+5
>> f(end-5)
ans =
     1
>> f(end+10)
ans =
    16

So IMO this behavior should be disallowed and should throw an error. But that may mean the parser would have to be smarter and examine the variable class before doing the replacement. Maybe a bug report or enhancement request to TMW is in order.

Michael 2018-3-23