我收到此错误:错误1:无法将类型'double [] []'隐式转换为'double []' [英] I got this error: Error 1: Cannot implicitly convert type 'double[][]' to 'double[]'

问题描述

我收到了以下代码:

I got this code:

double[][] ou = new double[h][w];

for(i=0;i<h;i++)
   for(int j=0;j<w;j++)
      ou[i][j] = sortie[z++];

return ou; 

我得到这2个错误:

And I get this 2 errors:

Error 1: Cannot implicitly convert type ''double[][]'' to ''double[]''

简称"return ou"

referred to "return ou"

Error 2: Invalid rank specifier: expected '','' or '']''

指的是"double [] [] ou = new double [h] [w];"




我解决了错误1，我的函数是double [] [] myFunction(...)

referred to "double[][] ou = new double[h][w];"




I solved Error 1, my function was double[][] myFunction(...)

推荐答案

已通读

Have a read through Arrays Tutorial (C#)[^] on MSDN.

Just as a hint, you have a jagged array in your code which is an array of arrays but I think what you are after is a multi-dimensional array.

很可能，您想要的是要做的是这样:

Most likely, what you want to do is this:

int h = 3; int w = 4; 
double[,] ou = new double[h, w];

你看到区别了吗?

如果这样做，您尝试写的内容可能很有意义:

Do you see the difference?

What you tried to write could make sense if you do this:

double[][] ou = new double[h][];

//now you have an array of h arrays of <code>double</code>,
//but each array of <code>double</code> is <code>null</code>;
//you can initialize each array of <code>double</code> separately (in loop for example);
//but let's look at couple of several elements:
ou[0] = new double[w];
uo[1] = new double [200];
uo[1] = new double [w * 2];

这样，double的每个内部数组可以具有不同的长度，因此是锯齿状数组".

现在，您可以了解Rod的建议，但是...为什么不简单地阅读C#手册(例如，从Microsoft帮助中获得)?

This way, each inner array of double can have different length, hence "jagged array".

Now, you can learn what Rod suggested but... why not simply reading C# manual (for example, from Microsoft help)?

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