C语言 在 C 程序中的函数中传递二维数组
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Passing a 2D array in a function in C program
提问by xyz
Below program (a toy program to pass around arrays to a function) doesn't compile. Please explain me, why is the compiler unable to compile(either because of technical reason or because of standard reason?)
下面的程序（一个将数组传递给函数的玩具程序）无法编译。请解释一下，为什么编译器无法编译（由于技术原因或标准原因？）
I will also look at some book explaining pointers/multi dimensional arrays(as I am shaky on these), but any off-the-shelf pointers here should be useful.
我还会看一些解释指针/多维数组的书（因为我对这些不太了解），但是这里的任何现成的指针都应该有用。
void print2(int ** array,int n, int m);
main()
{
int array[][4]={{1,2,3,4},{5,6,7,8}};
int array2[][2]={{1,2},{3,4},{5,6},{7,8}};
print2(array,2,4);
}
void print2(int ** array,int n,int m)
{
int i,j;
for(i=0;i<n;i++)
{
for(j=0;j<m;j++)
printf("%d ",array[i][j]);
printf("\n");
}
}
回答by cnicutar
This (as usual) is explained in the c faq. In a nutshell, an array decays to a pointer only once(after it decayed, the resulting pointer won't further decay).
这（像往常一样）在c 常见问题解答中进行了解释。简而言之，数组只衰减一次指针（衰减后，结果指针不会进一步衰减）。
An array of arrays (i.e. a two-dimensional array in C) decays into a pointer to an array, not a pointer to a pointer.
数组数组（即 C 中的二维数组）衰减为指向数组的指针，而不是指向指针的指针。
Easiest way to solve it:
最简单的解决方法：
int **array; /* and then malloc */
回答by Jens Gustedt
In C99, as a simple rule for functions that receive "variable length arrays" declare the bounds first:
在 C99 中，作为接收“可变长度数组”的函数的简单规则，首先声明边界：
void print2(int n, int m, int array[n][m]);
and then your function should just work as you'd expect.
然后你的函数应该像你期望的那样工作。
Edit: Generally you should have a look into the order in which the dimension are specified. (and me to :)
编辑：通常您应该查看指定维度的顺序。（和我：）
回答by Vishal Kanaujia
In your code the double pointer is not suitable to access a 2D array because it does not know its hop size i.e. number of columns. A 2D array is a contiguous allotted memory.
在您的代码中，双指针不适合访问二维数组，因为它不知道其跃点大小，即列数。二维数组是连续分配的内存。
The following typecast and 2D array pointer would solve the problem.
下面的类型转换和 2D 数组指针可以解决这个问题。
# define COLUMN_SIZE 4
void print2(int ** array,int n,int m)
{
// Create a pointer to a 2D array
int (*ptr)[COLUMN_SIZE];
ptr = int(*)[COLUMN_SIZE]array;
int i,j;
for(i = 0; i < n; i++)
{
for(j = 0; j < m; j++)
printf("%d ", ptr[i][j]);
printf("\n");
}
}
