您提供的代码很少有问题，我将一一介绍给您，

Your provided code have few issues i will explain them to you one by one ,

要将多维数组传递给函数，我将向您展示2种方法做到这一点

To pass multidimensional arrays to functions , I will show you 2 ways to do that

1）将多维数组作为一维数组传递
之所以可行，是因为我们知道数组如何在内存中表示，而我们对表示的了解正是使C / C ++中的指针如此强大的工具。
阅读此。

1) Passing multidimensional arrays as single dimensional arrays This works because we know how array is represented in the memory , and our this knowledge of representation is what makes pointers in C/C++ such a powerful tool . Read this answer to get a better picture of representation of array in memory.

元素（此处为ints）的填充方式如图片中的之字形线所描述。
因此，在我们的示例数组 int arr [5] [5] 中，
第一行的第二个元素（ arr [ 0] [1] ）可以通过 *（arr + 0 * 5 + 1）进行访问，就像 arr 给出数组的基地址，类似地， arr [4] [3] ） $ c> *（arr + 4 * 5 + 3），这里我们将行索引乘以5，因为每行有5个元素（即列数），有了这些知识，我们可以

Arrays are represented linearly and contiguously in memory , thus if define array as arr[5][5] , we are telling compilers that we need a memory block having sufficient space for storing 5*5 = 25 int data types . And it's also worth to know that arrays are represented in row major form , read this to learn more about row major form. .

Elements(here ints) are filled in the way as described by the zig-zag line in the picture . 
Thus in our example array int arr[5][5],
2nd element of the 1st row(arr[0][1]) can be accessed by *(arr+0*5+1) , as arr gives the base address of the array , similarly , 4th element of 5th row (arr[4][3]) can be accessed by *(arr+4*5+3) , here we are multiplying row index by 5 because each row have 5 elements(that is number of columns) , with this knowledge in mind we can write code to access array elements of a matrix in the following way 

在调用函数时，强制转换 arr 到（int *）是必要的，因为最初的arr类型是 int（*）[3] ，

void display(int *arr,int r,int c) { for(unsigned i=0;i<r;++i) { for(unsigned j=0;j<c;++j) { cout<<*(arr+i*c+j)<<ends; } cout<<endl; } } const unsigned ROW=3,COL=3; int main() { int arr[ROW][COL]={1,2,3, 4,5,6, 7,8,9 }; display((int *)arr,ROW,COL); }

2）将多维数组作为指针传递给函数参数中的数组。

While calling the function , casting arr to (int *) is necessary because originally arr type is int (*)[3] , that is pointer to an int array of 3 elements .

2） p>

2) Passing multidimensional array as pointer to an array in the function argument .

const unsigned ROW=3,COL=3;
void display(int (*arr)[COL],int r,int c)
{
    for(unsigned i=0;i<r;++i)
    {
        for(unsigned j=0;j<c;++j)
        {
            cout<<arr[i][j]<<ends;
        }
        cout<<endl;
    }
}
int main()
{
    int arr[ROW][COL]={1,2,3,
                       4,5,6,
                       7,8,9
                    };
    display(arr,ROW,COL);
}

在此，无需强制转换 arr

在您的代码中，您并没有在意数组索引超出范围，如果您的数组是 int arr [5] [5] 并且您尝试访问 arr [-1] [ 5] 或 arr [5] [3] ，结果将是不确定的，这就是代码崩溃到系统都会发生的一切

In your code you are not taking care of array index going out of bounds , if your array is int arr[5][5] and you try to access arr[-1][5] or arr[5][3] , the result will be undefined , that is anything can happen from your code just crashing to your system going up in flames(just a metaphor) .

记住这些事情，满足您需求的有效代码就是

Keeping these things in mind , a working code satisfying your needs is

int count(int *t, int r, int c)
{

    int i, j, result = 0;

    for (i = 0; i < r; i++)
    {

        for (j = 0; j < c; j++)
        {
            if(i!=r-1)
            {
                if(*(t+i*c+j)==*(t+(i+1)*c+j))
                    ++result;
                if(j!=r-1)
                {
                    if(*(t+i*c+j)==*(t+(i+1)*c+j+1))
                        ++result;
                }
                if(j!=0)
                {
                    if(*(t+i*c+j)==*(t+(i+1)*c+j-1))
                        ++result;
                }
            }
            if(j!=c-1)
            {
                if(*(t+i*c+j)==*(t+i*c+j+1))
                    ++result;
            }
        }
    }
    return result;
}
const unsigned ROW=3,COL=3;
int main()
{

    int arr[ROW][COL]={1,1,2,
                       1,2,3,
                       4,5,2
                    };
    cout<<count((int *)(arr),ROW,COL)<<endl;

}