为什么不能在MATLAB匿名函数中定义一个变量？ [英] Why cant I define a variable inside a MATLAB anonymous function?

问题描述

我一定错过了一件非常简单的事情，因为这看起来应该很难。

这段代码是正确的：

 清除所有
无论= @（x）交易（max（x），size（x））; 
 input = randn（1,1000）; 
 [ab] = whatever（输入）
  

然而，我真正想做的是像这样：

  clear all 
 whatever = @（x）deal（q = 3; q * max（ x），大小（x））; 
 input = randn（1,1000）; 
 [a b] =无论（输入）
  

为什么这会打破？我不能在函数内定义q？我想要使​​用匿名函数的全部原因是我可以在其中执行多行代码，然后返回答案。我假设匿名函数的最后一个语句是返回的，但是如何在其中定义变量？我不想在匿名函数的定义之前定义q。

谢谢。

解决方案

您不能在匿名函数中声明变量，因为它必须从表达式构造 ie ： handle = @（arglist）expr p如果您想要可读性， q 在函数外部，如下所示：

  q = 3; 
 whatever = @（x）deal（q * max（x），size（x））; 
  

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.

解决方案

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));

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