使用1malloc调用为2D矩阵分配内存

Question

We can allocate memory for 2d matrix using 1 malloc call as
int (*a)[5];
int i,j;

Answer 1

剪辑：整个故事。

前面我忽略了其他三种分配2D数组的方法。

动态二维数组方法1：

#define CCOLS 200

int (*m)[CCOLS] = malloc(cRows * sizeof(*m));
m[iRow][iCol] = n; // sets the item at iRow*CCOLS + iCol

...

free(m);

foo(int (*m)[CCOLS]) { ... }

foo(int m[][CCOLS]) { ... }

foo(int **m) { ... }

+---------+---------+---------+---------+     
| m[0][0] | m[0][1] | m[0][2] | m[0][3] |     
+---------+---------+---------+---------+     
| m[1][0] | m[1][1] | m[1][2] | m[1][3] |     
+---------+---------+---------+---------+     
| m[2][0] | m[2][1] | m[2][2] | m[2][3] |     
+---------+---------+---------+---------+     
| m[3][0] | m[3][1] | m[3][2] | m[3][3] |     
+---------+---------+---------+---------+     

int**M如下所示：

+----+        +----+----+----+----+----+      
|m[0]|  --->  |    |    |    |    |    |      
+----+        +----+----+----+----+----+      
|m[1]|  --->  |    |    |    |    |    |      
+----+        +----+----+----+----+----+      
|m[2]|  --->  |    |    |    |    |    |      
+----+        +----+----+----+----+----+      
|m[3]|  --->  |    |    |    |    |    |      
+----+        +----+----+----+----+----+      

动态2D数组方法2(C99，并非所有编译器都支持）：

这个和前面的一样，但是在编译时不需要知道维度。

int cCols, cRows, iCol, iRow;
... set cRows, cCols somehow, they could be passed in as parameters also ...
int (*m)[cCols] = malloc(cRows * sizeof(*m));
m[iRow][iCol] = n; // sets the item at iRow*cCols + iCol

...

free(m);

foo(int cCols, m[][cCols])  {}

foo(int cRows, int cCols, m[cRows][cCols])  {}

int cRows, cCols;
... set cRows, cCols somehow ...
int m[cRows][cCols];
m[iRow][iCol] = n; 

将其传递给函数的方式与方法2相同。

动态二维数组方法4：

这是很多人使用的指针数组方法。

int cCols = 10, cRows = 100, iRow;

// allocate:
// cCols*cRows*sizeof(int) = space for the data
// cRows*sizeof(int*) = space for the row ptrs
int **m = malloc(cCols*cRows*sizeof(int) + cRows*sizeof(int*));

// Now wire up the row pointers.  They take the first cRows*sizeof(int*) 
// part of the mem becasue that is what m[row] expects.
// we want each row pointer to have its own cCols sized array of ints.
// We will use the space after the row pointers for this.
// One way to calc where the space after the row pointers lies is to
// take the address of the nth + 1 element: &m[cRows].
// To get a row ptr, cast &m[cRows] as an int*, and add iRow*cCols to that.
for (iRow = 0; iRow < cRows; ++iRow)
    m[iRow] = (int*)&m[cRows] + iRow*cCols; 

// or 
for (p=(int*)&m[cRows] ; iRow = 0; iRow < cRows; ++iRow, p+=cCols)
    m[iRow] = p; 


// use it:
...
m[iRow][iCol] = 10;
...

// free it
free(m);