如何通过一个多维数组给函数没有在C ++中的内部尺寸？ [英] How to pass a multidimensional array to a function without inner dimension in c++?

问题描述

我不知道是否有一种方法说通过这个例如双MYARRAY [] [2] = {{0.1,0.8}，{0.4,0.6}} 到这样的功能无效MyFunction的（双myarray的[] []）; ，没有说这个无效MyFunction的（双myarray的[] [2] ）; 。我需要这个，因为我希望我的功能，能够处理例如与不同的内部尺寸数组：双MYARRAY [] [3] = {{0.1,0.8}，{0.4,0.6}，{0.3 ，0.9}} 。要尽量把它传递给一个函数，因为我知道我将不得不改变MyFunctions的参数，以的MyFunction（myarray的[] [3]）; 。从我读，我不知道这是可能的，因此，如果它不是，那么有没有这样做的其他方式？

I was wondering if there was a way to say pass this for example double MyArray[][2] = {{0.1,0.8},{0.4,0.6}} to a function like this void MyFunction(double myArray[][]);, without saying this void MyFunction(double myArray[][2]);. I need this because I want my function to be able to handle arrays with different inner dimensions for example: double MyArray[][3] = {{0.1,0.8},{0.4,0.6},{0.3,0.9}}. To pass this to a function as far as I know I would have to change MyFunctions's parameters to MyFunction(myArray[][3]);. From what I have read I don't know if this is possible, so if it is not then is there some other way of doing this?

推荐答案

您可以通过引用传递一个任意的二维数组，如果你可以改变的MyFunction 来：

You can pass an arbitrary 2d array by reference if you could change MyFunction to:

template<std::size_t N, std::size_t M>
void MyFunction(double (&myArray)[N][M]) {
  // ...
}

此方式，您也将有数组的大小。

This way you would also have dimensions of the array.

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