“在C#中使用MLapp时无效的被调用者”

问题描述 投票:0回答:3

使用MLApp.GetWorkspaceData函数时遇到一个奇怪的问题。我注意到当我执行以下操作时此功能可以正常工作:

matlab = new MLApp.MLAppClass();

object myObject;

matlab.GetWorkspaceData("myVariable", "base", out myObject);

但是如果我随后尝试将同一对象用作输出,则会收到“无效的被调用方”异常。此外,这还会产生相同的错误:

matlab = new MLApp.MLAppClass();

object myObject = new object();

matlab.GetWorkspaceData("myVariable", "base", out myObject);

这非常麻烦,因为我需要从Matlab到Visual Studio获取大量数据,并且实际上无法创建52K未初始化的变量并将其保留。有某种“未初始化”变量的方法吗?我在这里缺少一些概念吗?

c# matlab com com-interop matlab-deployment
3个回答
1
投票

如注释中的@wonko79所述,如果要重用out变量,应首先将其设置为null

这里是经过测试的示例calling MATLAB from C#

using System;

namespace CSharp_matlab_com
{
class Program
{
    static void Main(string[] args)
    {
        MLApp.MLAppClass matlab = new MLApp.MLAppClass();

        // create variables: a_0, a_1, ..., a_4
        for (int k = 0; k < 5; k++) {
            matlab.Execute(string.Format("a_{0} = rand(2);", k));
        }

        // retrieve variables from MATLAB and print their contents
        object a;
        for (int k = 0; k < 5; k++) {
            // current variable name
            string varname = string.Format("a_{0}", k);

            // get data array
            a = null;    // without this line, an exception is thrown!
            matlab.GetWorkspaceData(varname, "base", out a);

            // print contents
            var arr = (double[,]) a;
            Console.WriteLine("\nndims(a) = {0}, numel(a) = {1}", arr.Rank, arr.Length);
            for (int i = 0; i < arr.GetLength(0); i++) {
                for (int j = 0; j < arr.GetLength(1); j++) {
                    Console.WriteLine("{0}[{1},{2}] = {3}", varname, i, j, arr[i,j]);
                }
            }
        }
    }
}
}

输出:

ndims(a) = 2, numel(a) = 4
a_0[0,0] = 0.251806122472313
a_0[0,1] = 0.617090884393223
a_0[1,0] = 0.290440664276979
a_0[1,1] = 0.265280909810029

...

ndims(a) = 2, numel(a) = 4
a_4[0,0] = 0.425259320214135
a_4[0,1] = 0.16148474431175
a_4[1,0] = 0.312718886820616
a_4[1,1] = 0.178766186752368

0
投票

您可以为GetWorkspaceData方法创建一个包装器,如下面的示例所示:

public object GetData(string name)
{
    object data;
    mlApp.GetWorkspaceData(name, "base", out data);

    return data;
}

或更有用的是通用包装:

public T GetData<T>(string name)
{
    object data;
    mlApp.GetWorkspaceData(name, "base", out data);

    if (data == null)
        return default(T);

    if (data is T)
        return (T)data;
    else
        throw new InvalidCastException($"The variable '{name}', of type '{data.GetType().Name}' cannot be casted to type '{typeof(T).Name}'.");
}

0
投票

解决方案是将输出对象设置为null

我找到了here

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