Question

//header file

CGPoint **positions;

//Implementation file
int rows = 10;
int columns = 6;

positions = malloc(rows * sizeof(CGPoint));

for(int i=0; i<rows; i++){
    positions[i] = malloc(columns * sizeof(CGPoint));
}

positions[0][0] = CGPointMake(682, 0);
positions[0][1] = CGPointMake(682, 336);
positions[0][2] = CGPointMake(0, 0);
positions[0][3] = CGPointMake(-341, 336);
positions[0][4] = CGPointMake(0, 336);
positions[0][5] = CGPointMake(341, 0);

positions[1][0] = CGPointMake(341, 0);
positions[1][1] = CGPointMake(341, 336);
positions[1][2] = CGPointMake(682, 336);
positions[1][3] = CGPointMake(-341, 336);
positions[1][4] = CGPointMake(0, 336);
positions[1][5] = CGPointMake(0, 0);

//and so on..

我需要帮助编写以下函数，这将返回像这样的随机第二维位置。返回完整的子阵列位置[0]或位置[1]，

- (CGPoint *)point {
    return positions[arc4random() % rows];
}

Answer 1

您应该分配指针大小而不是结构大小。
~~CGPointMake(x, y)在堆栈中创建结构而不是堆。~~感谢@Bavarious指出这不是真的。见下面的评论。：）

执行您想要的代码：

unsigned int row = 10;
unsigned int column = 10;
CGPoint **points = malloc(row * sizeof(CGPoint *));
for (int r = 0; r < row; ++r) {
    points[r] = malloc(column * sizeof(CGPoint));
    for (int c = 0; c < column; ++c) {
        points[r][c].x = 0.0;
        points[r][c].y = 0.0;
    }
}

警告：当您不再需要时，您必须记住释放2D阵列。一种方法是将其包装在Obj-C包装器中，并在init和dealloc中执行此操作。

返回动态生成的2D数组的引用

1 个答案: