0

可能重复：
创建指向二维数组的指针

当我调用函数 func4() 和 func5() 时，出现以下错误：

func4() 错误：无法将参数 '1' 的 'short int (*)[3]' 转换为 'short int**' 到 'int func4(short int**)'| func5() 错误：无法将参数 '1' 的 'short int (*)[3]' 转换为 'short int**' 到 'int func5(short int**)'|

如何更正调用函数 func4() 和 func5() 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;
}
4

3 回答 3

1

You can't send a two dimensional array to function without specifying the length or size of second dimension of the array. This is what causing the error.

Try this:

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 this that if something can be done easily and cleanly through one way, don't try to do it through a dirty and difficult way as it will confuse yourself when you are revising or updating your code in the future.

于 2012-12-31T10:31:59.090 回答
1

1.它们是一样的：

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

并且与short **不兼容short (*)[3]，因为它们指向的类型不同。short **指向short *whileshort (*)[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.我认为第三个功能是最好的！它使用了我在POINTERS ON C中读到的一个名为“扁平化数组”的技巧。

方法#3（使用单个指针，数组被“扁平化”）================================== =========================== 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，因此您可以将它们添加到正式参数列表中。

手动数组索引可能会减慢执行速度。

于 2012-12-31T11:40:32.563 回答
0

问题在于 和 的矩阵定义不func4()正确func5()。

您将其定义为short mat[3][3]，但在这种情况下，您实际上并未分配指向矩阵行的指针数组。您将获得一个指向一个连续内存块的指针。

如果要将参数矩阵传递为short int**，则应将其定义如下：

#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;
    }
}
于 2012-12-31T10:21:49.473 回答