将一个二维数组作为参数传递给函数

If I dont know the size of both the dimensions of array and want to print a matrix using the following code

如果我不知道数组的两个维度的大小，想用下面的代码打印一个矩阵

    
    void printAnyMatrix(int (*A)[], int size_A, int size_B)
    {
       for (int i = 0; i<=size_A; i++)
       {
           for (int j = 0; j<=size_B; j++)
               printf("%d ", A[i][j]);
           printf("\n");
       }
       printf("\n");
    }

Compiler gives

error cannot convert ‘int (*)[(((unsigned int)((int)size_B)) + 1)]’ to ‘int ()[]’ for argument ‘1’ to ‘void printAnyMatrix(int ()[], int, int)

错误无法将' int(*)[((((((无符号int)size_B)) + 1)] '转换为' int ()[] ' for arguments ' 1 '转换为' void printAnyMatrix(int ()[]， int, int)

3 个解决方案

#1

Use template feature for such problems:

对此类问题使用模板特性:

template<typename T, unsigned int size_A, unsigned int size_B>
void printAnyMatrix(T  (&Arr)[size_A][size_B])
{       // any type^^  ^^^ pass by reference        
}

Now you can pass any 2D array to this function and the size will be automatically deduced in the form of size_A and size_B.

现在您可以将任何2D数组传递给这个函数，大小将自动以size_A和size_B的形式推断出来。

Examples:

例子:

int ai[3][9];
printAnyMatrix(ai);
...
double ad[18][18];
printAnyMatrix(ad);

#2

Simplify the signature: a pointer is simpler to read by humans.

简化签名:指针更容易被人类读取。

You also have an error in the loops: it's less then, not less or equal.

在循环中也有一个错误:它小于，而不是小于或等于。

void printAnyMatrix(int *A, int size_A, int size_B)<br />
{
   for (int i = 0; i<size_A; i++)
   {
       for (int j = 0; j<size_B; j++)
           printf("%d ", A[i*size_B + j]);
       printf("\n");
   }
   printf("\n");
}

#3

If you want to print any matrix, maybe you will need to print both statically and dynamically allocated matrices. Function for statically-allocated matrices will look like the following

如果要打印任何矩阵，可能需要同时打印静态和动态分配的矩阵。静态分配矩阵的函数将如下所示

template <class T, int size_A, int size_B>
void printAnyMatrix(T (&A)[size_A][size_B])
{
   for (int i = 0; i < size_A; i++)
   {
       for (int j = 0; j < size_B; j++)
           std::cout<<A[i][j]<<' ';
       std::cout<<'\n';
   }
   std::cout<<std::endl;
}

Function for dynamically-allocated matrices:

函数动态矩阵:

template <class T>
void printAnyMatrix(T **A, int size_A, int size_B)
{
   for (int i = 0; i < size_A; i++)
   {
       for (int j = 0; j < size_B; j++)
           cout<<A[i][j]<<' ';
       cout<<'\n';
   }
   std::cout<<std::endl;
}

You can have both of them in the same translation unit, compiler will pick the one that fits to your matrix. (note: for statically-allocated matrices you need one parameter, while for dynamically-allocated - three)

你可以把它们放在同一个翻译单元中，编译器会选择适合你的矩阵的那个。(注:静态分配矩阵需要一个参数，动态分配矩阵需要三个参数)

Please consider using cout instead of printf if you work in C++. It is overloaded for all basic types, while printf needs you to declare the type explicitly.

如果您使用c++，请考虑使用cout而不是printf。它对所有基本类型都是重载的，而printf需要您显式地声明类型。

If you still have compiler errors, please show the declaration of your matrix.

如果您仍然有编译错误，请显示您的矩阵的声明。

#1