2暗数组和双指针

Question

可能重复： Create a pointer to two-dimensional array

当我调用函数func4()和func5()时，会得到以下错误：

func4() error:不能将参数‘1’的‘短int (*)3’转换为‘int **’到‘int func4(短int**)’区func5()错误:不能将参数‘1’的‘short int(*)3’转换为‘int**’到‘int func5(短int**)’。

如何纠正调用函数func4()和func5()时的错误？这是我的代码：

#include <cstdio>

int func1(short mat[][3]);
int func2(short (*mat)[3]);
int func3(short *mat);
int func4(short **mat);
int func5(short *mat[3]);

int main()
{

short mat[3][3],i,j;

for(i = 0 ; i < 3 ; i++)
    for(j = 0 ; j < 3 ; j++)
    {
        mat[i][j] = i*10 + j;
    }

printf(" Initialized data to: ");
for(i = 0 ; i < 3 ; i++)
{
    printf("\n");
    for(j = 0 ; j < 3 ; j++)
    {
        printf("%5.2d", mat[i][j]);
    }
}

printf("\n");

func1(mat);
func2(mat);
func3(&mat[0][0]);
func4(mat); //error: cannot convert ‘short int (*)[3]’ to 
            //‘short int**’ for argument ‘1’ to       ‘int func4(short int**)’|
func5(mat); //error: cannot convert ‘short int (*)[3]’ to 
            //‘short int**’ for argument ‘1’ to ‘int func5(short int**)’|

return 0;
}



/*
Method #1 (No tricks, just an array with empty first dimension)
===============================================================
You don't have to specify the first dimension!
*/

int func1(short mat[][3])
{
register short i, j;

printf(" Declare as matrix, explicitly specify second dimension: ");
for(i = 0 ; i < 3 ; i++)
{
    printf("\n");
    for(j = 0 ; j < 3 ; j++)
    {
        printf("%5.2d", mat[i][j]);
    }
}
printf("\n");

return 0;
}

/*
Method #2 (pointer to array, second dimension is explicitly specified)
======================================================================
*/

int func2(short (*mat)[3])
{
register short i, j;

printf(" Declare as pointer to column, explicitly specify 2nd dim: ");
for(i = 0 ; i < 3 ; i++)
{
    printf("\n");
    for(j = 0 ; j < 3 ; j++)
    {
        printf("%5.2d", mat[i][j]);
    }
}
printf("\n");

return 0;
}

/*
Method #3 (Using a single pointer, the array is "flattened")
============================================================
With this method you can create general-purpose routines.
The dimensions doesn't appear in any declaration, so you
can add them to the formal argument list.

The manual array indexing will probably slow down execution.
*/

int func3(short *mat)
{
register short i, j;

printf(" Declare as single-pointer, manual offset computation: ");
for(i = 0 ; i < 3 ; i++)
{
    printf("\n");
    for(j = 0 ; j < 3 ; j++)
    {
        printf("%5.2d", *(mat + 3*i + j));
    }
}
printf("\n");

return 0;
}

/*
Method #4 (double pointer, using an auxiliary array of pointers)
================================================================
With this method you can create general-purpose routines,
if you allocate "index" at run-time.

Add the dimensions to the formal argument list.
*/

int func4(short **mat)
{
short    i, j, *index[3];

for (i = 0 ; i < 3 ; i++)
    index[i] = (short *)mat + 3*i;

printf(" Declare as double-pointer, use auxiliary pointer array: ");
for(i = 0 ; i < 3 ; i++)
{
    printf("\n");
    for(j = 0 ; j < 3 ; j++)
    {
        printf("%5.2d", index[i][j]);
    }
}
printf("\n");

return 0;
}

/*
Method #5 (single pointer, using an auxiliary array of pointers)
================================================================
*/

int func5(short *mat[3])
{
short i, j, *index[3];
for (i = 0 ; i < 3 ; i++)
    index[i] = (short *)mat + 3*i;

printf(" Declare as single-pointer, use auxiliary pointer array: ");
for(i = 0 ; i < 3 ; i++)
{
    printf("\n");
    for(j = 0 ; j < 3 ; j++)
    {
        printf("%5.2d", index[i][j]);
    }
}
printf("\n");
return 0;
}

Answer 1

您的can't不指定数组的length or size of second dimension就发送一个二维数组来工作。这就是导致错误的原因。

试试这个：

int func4(short mat[][3])
{
short    i, j, *index[3];

for (i = 0 ; i < 3 ; i++)
index[i] = (short *)mat + 3*i;

printf(" Declare as double-pointer, use auxiliary pointer array: ");
for(i = 0 ; i < 3 ; i++)
{
printf("\n");
for(j = 0 ; j < 3 ; j++)
{
    printf("%5.2d", index[i][j]);
}
}
printf("\n");

return 0;
}


int func5(short mat[][3])
{
short i, j, *index[3];
for (i = 0 ; i < 3 ; i++)
index[i] = (short *)mat + 3*i;

printf(" Declare as single-pointer, use auxiliary pointer array: ");
for(i = 0 ; i < 3 ; i++)
{
printf("\n");
for(j = 0 ; j < 3 ; j++)
{
    printf("%5.2d", index[i][j]);
}
}
printf("\n");
return 0;

}

Remember --如果某件事情可以通过一种方式轻松而干净地完成，那么不要试图通过一种肮脏而困难的方式来完成它，因为将来在修改或更新代码时，它会使自己感到困惑。

Answer 2

1.它们是相同的：

int func(short **mat);
int func(short *mat[]);
int func(short *mat[3]);

而且short **与short (*)[3]不兼容，因为它们指向的类型是不同的。short **指向short *，short (*)[3]指向short[3]。

2.也许你可以试试这个：

int funcMy(void *mat)   // OK! -added by Justme0 2012/12/31
{
    short i, j, *index[3];
    for (i = 0 ; i < 3 ; i++)
        index[i] = (short *)mat + 3*i;

    printf(" Declare as (void *) pointer, use auxiliary pointer array: ");
    for(i = 0 ; i < 3 ; i++)
    {
        printf("\n");
        for(j = 0 ; j < 3 ; j++)
        {
            printf("%5.2d", index[i][j]);
        }
    }
    printf("\n");
    return 0;
}

3.我认为第三场比赛是最好的！它使用了一个名为“扁平数组”的技巧，我在C上的指针中看到了这个技巧。

方法3(使用单个指针，数组是“扁平的”) ============================================================，使用此方法您可以创建通用例程。维度不会出现在任何声明中，所以可以将它们添加到正式的参数列表中。 手动数组索引可能会减慢执行速度。

Answer 3

问题在于不正确地定义func4()和func5()的矩阵。

您将其定义为short mat[3][3]，但在本例中，实际上不为矩阵行分配指针数组。您将得到一个连续内存块的一个指针。

如果要将参数矩阵传递为short int**，则应将其定义为：

#include <stdlib.h>

short int** mat;

for(int i = 0; i < 3; i++) {
    mat[i] = (short int*)malloc (3*sizeof(short int));
    for(int j = 0; j < 3; i++) {
         mat[i][j] = i*10 + j;
    }
}