如何将二维数组的引用传递给函数? [英] How do I pass a reference to a two-dimensional array to a function?

问题描述

我正在尝试将二维数组的引用传递给 C++ 中的函数.我知道编译时两个维度的大小.这是我现在所拥有的:

I am trying to pass a reference to a two-dimensional array to a function in C++. I know the size of both dimensions at compile time. Here is what I have right now:

const int board_width = 80;
const int board_height = 80;
void do_something(int[board_width][board_height]& array);  //function prototype

但这行不通.我从 g++ 得到这个错误:

But this doesn't work. I get this error from g++:

error: expected ‘,’ or ‘...’ before ‘*’ token

这个错误是什么意思，我该如何解决?

What does this error mean, and how can I fix it?

推荐答案

如果你知道编译时的大小，这样做:

If you know the size at compile time, this will do it:

//function prototype
void do_something(int (&array)[board_width][board_height]);

一起做

void do_something(int array[board_width][board_height]);

实际上会传递一个指向二维数组的第一个子数组的指针(board_width"被完全忽略，就像当你有 int array[] 接受一个指针)，这可能不是您想要的(因为您明确要求提供参考).因此，使用引用进行操作，在参数 sizeof array 上使用 sizeof 将产生 sizeof(int[board_width][board_height])(就像你会在参数本身)同时使用第二种方法(将参数声明为数组，从而使编译器将其转换为指针)将产生 sizeof(int(*)[board_height])，因此仅指针的大小.

Will actually pass a pointer to the first sub-array of the two dimensional array ("board_width" is completely ignored, as with the degenerate case of having only one dimension when you have int array[] accepting a pointer), which is probably not what you want (because you explicitly asked for a reference). Thus, doing it with the reference, using sizeof on the parameter sizeof array will yield sizeof(int[board_width][board_height]) (as if you would do it on the argument itself) while doing it with the second method (declaring the parameter as array, thus making the compiler transform it to a pointer) will yield sizeof(int(*)[board_height]), thus merely the sizeof of a pointer.

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