如何使用c型数组作为伊娃？ [英] How to use a c-type array as an ivar?

问题描述

我需要整数在我的程序一个老式的2维数组，但无论我如何努力将其声明为伊娃，然后用@财产/ @合成，我得到一个编译器的投诉或其他。

I need a good old-fashioned 2-dimensional array of integers in my program, but no matter how I try to declare it as an ivar, and then use @property/@synthesize, I get one compiler complaint or another.

我宣布

int spotLocations[10] [10]

作为伊娃。

这作品不多，但随后的@财产/ @合成过程中从来没有通过审核。

That much works, but then the @property/@synthesize process never passes muster.

推荐答案

您不能做到这一点。阵列variabless永远不能在C，这意味着你可以永远声明返回一个数组的函数左值，因为这将是不可能给函数的结果分配给一个数组变量（因为它不能是一个左值）。

You can't do this. Array variabless can never be lvalues in C, which means you can never declare a function that returns an array, because it would be impossible to assign the result of the function to an array variable (since it can't be an lvalue).

属性是声明返回类型的函数只是一个简便方法。由于函数永远不能返回数组，你永远不能声明一个属性，是一个数组。

Properties are just a shorthand way of declaring a function that returns a type. Since functions can never return arrays, you can never declare a property that is an array.

如果您确实需要移动矩阵左右这个样子，你可以包装在一个结构，它可以是左值：

If you absolutely need to move matrices around like this, you could wrap it in a struct, which can be lvalues:

typedef struct {
  int value[10][10];
} matrix;

...

@property matrix spotLocations;

当然，访问的地点是一个小更令人费解，你必须使用

Of course, accessing the locations is a little more convoluted, you have to use

spotLocations.value[x][y]

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