如何编写一个可以同时采用动态/静态分配的2D数组的c函数？ [重复]

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• Why do I need to use type** to point to type*? 3 answers

为什么我需要使用类型**指向类型*? 3个答案

I have a function that supposed to take 2D array as an argument, my code looks like this --

我有一个假设将2D数组作为参数的函数,我的代码看起来像这样 -

#include <stdio.h>
#include <stdlib.h>

void func(double**, int);

int main()
{
    double m[3][3] = {{1, 1, 1}, {2, 2, 2}, {3, 3, 3}};
    func(m, 3);
}

void func(double **m, int dim)
{
    int i, j ;
    for(i = 0 ; i < dim ; i++)
    {
        for(j = 0 ; j < dim ; j++)
            printf("%0.2f ", m[i][j]);
        printf("\n");
    }
}

Then the compiler says --

然后编译器说 -

test.c: In function ‘main’:
test.c:9:2: warning: passing argument 1 of ‘func’ from incompatible pointer type [enabled by default]
  func(m, 3);
  ^
test.c:4:6: note: expected ‘double **’ but argument is of type ‘double (*)[3]’
 void func(double**, int);
      ^

But when I say --

但是当我说 -

int main()
{
    int i, j;
    double m[3][3] = {{1, 1, 1}, {2, 2, 2}, {3, 3, 3}};
    double **m1 ;
    m1 = (double**)malloc(sizeof(double*) * 3);
    for(i = 0 ; i < 3 ; i++)
    {
        m1[i] = (double*)malloc(sizeof(double) * 3);
        for(j = 0 ; j < 3 ; j++)
            m1[i][j] = m[i][j] ;
    }
    func(m1, 3);
    for(i = 0 ; i < 3 ; i++) free(m1[i]);
    free(m1);
}

It compiles and runs.

它编译并运行。

Is there any way I can make func() to take both the statically/dynamically defined 2D array ? I am confused since I am passing the pointer m, why it is not correct for the first case ?

有什么方法可以使func()同时采用静态/动态定义的2D数组?我很困惑,因为我传递指针m,为什么第一种情况不正确?

does it mean that I need to write two separate functions for two different types of arguments?

这是否意味着我需要为两种不同类型的参数编写两个单独的函数?

1 个解决方案

#1

Your dynamic allocation for 2d array is not correct. Use it as:

您对2d数组的动态分配不正确。用它作为:

double (*m1)[3] = malloc(sizeof(double[3][3]));

And then it will work.

然后它会工作。

Moreover change the function prototype to:

而且将函数原型更改为:

void func(double m[][3], int dim)

Another way is to use a 1-D array of size w * h instead of 2-D array.

另一种方法是使用大小为w * h的1-D阵列而不是2-D阵列。

Working example

---

From comment of @TheParamagneticCroissant c99 onwards, you can also VLA's and make both your dimensions variable. (You'll need to allocate the 2D array properly though)

从@TheParamagneticCroissant c99的评论开始,您还可以使用VLA并使您的维度变量。 (你需要正确分配2D数组)

Change function signature to:

将功能签名更改为:

void func(int dim, double[dim][dim]);  /* Second arg is VLA whose dimension is first arg */
/* dim argument must come before the array argument */

Working example