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

我正在尝试编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.

由于我没有说清楚，传递给这个函数的数组不是动态分配的.大小在编译时已知.

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

我正在使用 3x2 阵列对其进行测试.这是它的功能:

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 << "
array1[" << i << "][" << j << "] = "
                 << setfill('0') << setw(2) << array1[i][j]
                 << "	array2[" << i << "][" << j << "] = "
                 << setfill('0') << setw(2) << array2[i][j];
        }
    }
}

显然，这仅在我知道i"的大小为 3 时才有效(在这种情况下).然而，理想情况下，无论第一维的大小如何，我都希望该函数能够工作.

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() 函数来做到这一点，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运算.

... 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，它会返回一个值 4.我可以使用指针表示法来取消对数组的引用:

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);

... 但这实际上返回了 8 的值.这很奇怪，因为总大小应该是行(= 3)* 列(= 2)* sizeof(int)(= 4)，或 24.事实上，这就是我在函数之外使用 sizeof(*array1) 时的结果.

... 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.

有谁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 是计算的编译时间；sizeof 在确定 C/C++ 中对象的动态大小时从来没有用.您(您自己，程序员)总是可以通过查看代码和头文件来计算 sizeof(x)，因为 sizeof 计算用于表示对象的字节数.sizeof(*array1) 将始终为 8，因为 array1[i] 是两个 ints 和 4==sizeof(int).当您声明 int array1[][2] 时，这等效于 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，因为在您的机器上需要 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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