C ++：获取传递给函数的多维数组的行大小 [英] C++: getting the row size of a multidimensional array passed to a function

问题描述

我想写一个函数来打印多维数组的内容。我知道列的大小，但不知道行的大​​小。

I'm trying to write a function that will print out the contents of a multidimensional array. I know the size of the columns, but not the size of the rows.

编辑：因为我没有这么清楚，传递给这个函数的数组不是动态分配。在编译时已知大小。

Since I didn't make this clear, the arrays passed to this function are NOT dynamically allocated. The sizes are known at compile time.

我使用3x2阵列测试它。这里是它的功能：

I am testing it using a 3x2 array. Here is the function as it stands:

void printArrays(int array1[][2], int array2[][2]) {
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 2; j++) {
            cout << "\narray1[" << i << "][" << j << "] = "
                 << setfill('0') << setw(2) << array1[i][j]
                 << "\tarray2[" << i << "][" << j << "] = "
                 << setfill('0') << setw(2) << array2[i][j];
        }
    }
}

显然，我知道i的大小是3（在这种情况下）。但是，理想情况下，无论第一维的大小如何，我都希望该函数可以工作。

Obviously, this only works if I know the size of "i" is 3 (it is in this case). Ideally, however, I would like the function to work no matter what the size of the first dimension.

我想我可以使用sizeof（）函数，例如

I thought I would be able to do this using the sizeof() function, e.g.

int size = sizeof(array1);

...并从那里做一些数学。

... and do some math from there.

这是奇怪的部分。如果我在数组中使用sizeof（）函数，它返回值4.我可以使用指针表示法来解引用数组：

Here's the odd part. If I use the sizeof() function inside the array, it returns a value of 4. I can use pointer notation to dereference the array:

int size = sizeof(*array1);

...但实际上返回的值为8.这很奇怪，是行（这是= 3）*列（= 2）* sizeof（int）（= 4）或24.而且，这是结果，当我使用sizeof（* array1） >的功能。

... but this actually returns a value of 8. This is odd, because the total size should be rows(which = 3) * columns(= 2) * sizeof(int)(= 4), or 24. And, indeed, this is the result, when I use sizeof(*array1) outside of the function.

有没有人知道这里发生了什么？更重要的是，有人有解决方案吗？

Does anyone know what is going on here? More importantly, does anyone have a solution?

推荐答案

答案是，你不能这样做。您必须将行数作为参数传递给函数，或者使用STL容器，例如 std :: vector 或 std :: array 。

The answer is that you can not do this. You must pass the number of rows as an argument to the function, or use an STL container such as std::vector or std::array.

sizeof sizeof 永远不会用于确定C / C ++中对象的动态大小。你自己，程序员总是可以从 sizeof 查看代码和头文件计算 sizeof（x）计算用于表示对象的字节数。 array1 [i] 是两个的数组，因此 sizeof（* array1） > ints 和 4 == sizeof（int）。当你声明 int array1 [] [2] 这相当于 int * array1 [2] 。也就是说， array1 是一个指向两个整数数组的指针。 sizeof（array1）因此是4个字节，因为在你的机器上需要4个字节来表示一个指针。

sizeof is computed compile time; sizeof is never useful in determining dynamic size of objects in C/C++. You (yourself, the programmer) can always calculate sizeof(x) just from looking at code and header files since sizeof counts the number of bytes used to represent the object. sizeof(*array1) will always be 8 since array1[i] is an array of two ints and 4==sizeof(int). When you declare int array1[][2] this is equivalent to int *array1[2]. That is, array1 is a pointer to arrays of two integers. sizeof(array1) is therefore 4 bytes, since it takes 4 bytes on your machine to represent a pointer.

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