问题
I must be missing something really simple because this doesnt seem like it should be this hard.
This code is correct:
clear all
whatever = @(x) deal(max(x), size(x));
input = randn(1,1000);
[a b] = whatever(input)
However, what I really want to do is something like this:
clear all
whatever = @(x) deal(q = 3; q*max(x), size(x));
input = randn(1,1000);
[a b] = whatever(input)
Why does this break? I cant define q inside the function?? The whole reason I want to use anonymous functions is so that I can actually do multiple lines of code within them, and then return an answer. I suppose the last statement of an anonymous function is what is returned, but how do I define variables within them? I dont want to define q before the definition of the anonymous function.
Thanks.
回答1:
You cannot declare variables inside an anonymous function, because it must be constructed from an expression, i.e.: handle = @(arglist)expr
If you want readability, define q outside the function, like this:
q = 3;
whatever = @(x) deal(q * max(x), size(x));
回答2:
You don't. Anonymous functions have only a single statement. You use subfunctions for that (not a nested function, those are sick things with strange scoping rules).
function whatever = not_anonymous (x)
% your code here
end
If you need to pass function handles, you can just use @not_anonymous.
回答3:
What do you think of following construct:
tmpfun = @(x,q) deal...
whatever = @(x) tmpfun(x,3)
回答4:
I'm pretty sure deal can't take in multiple commands. Multiple parameters, sure, but you're trying to pass in commands. Would this work?
whatever = @(x) q=3; deal(q*max(x), size(x));
Also, why wouldn't you just have this?
whatever = @(x) deal(3*max(x), size(x));
If you're going to define it within the function, you might as well just put the actual value there, if you can't get anything else to work.
来源：https://stackoverflow.com/questions/12229394/why-cant-i-define-a-variable-inside-a-matlab-anonymous-function