Question

因为我似乎无法编辑我的旧问题（Matlab - Function taking no arguments but not static），所以这里又是：

我正在尝试实施以下内容：

classdef asset
    properties
        name
        values
    end    

    methods

        function AS = asset(name, values)
            AS.name = name;
            AS.values = values;
        end

        function out = somefunction1
            ret = somefunction2(asset.values);
            out = mean(ret);
            return
        end

        function rets = somefunction2(vals)
            n = length(vals);
            rets = zeros(1,n-1);
            for i=1:(n-1)
                rets(i) = vals(i)/vals(i+1);
            end
            return
        end
    end
end

但我得到的错误是somefunction1应该是静态的。但如果它是静态的，那么它就不能再访问这些属性了。我该如何解决这个问题？

基本上我希望能够写出这样的东西：

AS = asset('testname',[1 2 3 4 5]);
output = AS.somefunction1();

而非写作

AS = asset('testname',[1 2 3 4 5]);
output = AS.somefunction1(AS);

Answer 1

要访问方法中对象的属性，需要将该对象作为参数传递给方法。如果您不需要特定对象来执行某个功能，那么请将其设置为静态（属于该类，但不对特定对象进行操作）。

所以，比较原始代码：

methods

    % ...

    function out = somefunction1
        ret = somefunction2(asset.values);
        out = mean(ret);
        return
    end;

    function rets = somefunction2(vals)
        n = length(vals);
        rets = zeros(1,n-1);
        for i=1:(n-1)
            rets(i) = vals(i)/vals(i+1);
        end
        return
    end
end

使用正确的代码：

methods

    % ...

    % this function needs an object to get the data from,
    % so it's not static, and has the object as parameter.

    function out = somefunction1(obj)
        ret = asset.somefunction2(obj.values);
        out = mean(ret);
    end;
end;

methods(Static)
    % this function doesn't depend on a specific object,
    % so it's static.

    function rets = somefunction2(vals)
        n = length(vals);
        rets = zeros(1,n-1);
        for i=1:(n-1)
            rets(i) = vals(i)/vals(i+1);
        end;
    end;
end;

要调用该方法，您确实要写（请测试）：

AS = asset('testname',[1 2 3 4 5]);
output = AS.somefunction1();

因为，在MATLAB中，99.99％的案例相当于：

AS = asset('testname',[1 2 3 4 5]);
output = somefunction1(AS);

当您为类重写subsref时，或者传递给方法的对象不是参数列表中的第一个（但这些是您不应该关注的情况）时，会出现差异现在，直到你澄清MATLAB类语义）。

Matlab - 函数在类中不带参数

1 个答案: