传递给函数的多维数组的指针类型是什么? [英] What is the pointer type of a multidimensional array when passed to a function?

问题描述

我在大学课堂上正在学习C和指针，我认为除了多维数组和指针之间的相似性以外，我对这个概念已经有了很好的了解.

I'm learning about C and pointers in my university class, and I think I've got a pretty good grasp on the concept except for the similarity between multidimensional arrays and pointers.

我认为，由于所有数组(甚至多维数组)都存储在连续的内存中，因此您可以安全地将其转换为int*(假设给定的数组为int[].)但是，我的教授说，数字定义中的星数取决于数组中维数的数量.因此，int[]将成为int*，int[][]将成为int**，依此类推.

I thought that since all arrays (even multidimensional) ones are stored in continuous memory, you could safely convert it to a int* (assuming the given array is an int[].) However, my professor said that the number of stars in the definition depends on the number of dimensions in the array. So, an int[] would become an int*, a int[][] would become an int**, etc.

所以我写了一个小程序来测试这个:

So I wrote a small program to test this:

void foo(int** ptr)
{

}

void bar(int* ptr)
{

}

int main()
{
    int arr[3][4];

    foo(arr);
    bar(arr);
}

令我惊讶的是，编译器在两个函数调用上均发出了警告.

To my surprise, the compiler issued warnings on both function calls.

main.c:22:9: warning: incompatible pointer types passing 'int [3][4]' to parameter of type
      'int **' [-Wincompatible-pointer-types]
    foo(arr);
        ^~~
main.c:8:16: note: passing argument to parameter 'ptr' here
void foo(int** ptr)
         ^
main.c:23:9: warning: incompatible pointer types passing 'int [3][4]' to parameter of type
      'int *' [-Wincompatible-pointer-types]
    bar(arr);
        ^~~
main.c:13:15: note: passing argument to parameter 'ptr' here
void bar(int* ptr)
         ^
2 warnings generated.

这是怎么回事?您如何更改函数调用，以便其中一个可以接受数组?

What's going on here? How could you change the function calls so one of them would accept the array?

这不是

This is not a duplicate of How to pass a multidimensional array to a function in C and C++ because I'm asking how to change the function calls themselves, not the function signatures.

推荐答案

数组和指针并非完全可以自由转换.为了不发出警告，您需要使用void foo(int (*ptr)[4])之类的签名.将类似的数组传递给函数时，只能用*替换第一个[].

Arrays and pointers aren't entirely freely convertible. To not get a warning, you'd need a signature like void foo(int (*ptr)[4]). You can only replace the first [] with a * when passing an array like that to a function.

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