Question

我想将矩阵（3x3数组）发送到另一个函数，以便将它乘以某些东西。但我希望这个矩阵的内容是有条件的。

APPROCH＃1：

 typedef float Mat3[3][3];

 void rotate(Point *p, float theta){

int choice = ((int)theta/90)%4;

switch(choice){
    case 0: matMult(p, {{1.0,0.0,0.0}, {0.0,1.0,0.0}, {0.0,0.0,1.0}}); break;
    case 1: matMult(p, {{0.0,-1.0,0.0}, {1.0,0.0,0.0}, {0.0,0.0,1.0}}); break;
    case 2: matMult(p, {{-1.0,0.0,0.0}, {0.0,-1.0,0.0}, {0.0,0.0,1.0}}); break;
    case 3: matMult(p, {{0.0,1.0,0.0}, {-1.0,0.0,0.0}, {0.0,0.0,1.0}}); break;
    default: matMult(p, {{1.0,0.0,0.0}, {0.0,1.0,0.0}, {0.0,0.0,1.0}}); break;
}
 }

APPROACH＃2：

void rotate(Point *p, float theta){

int choice = ((int)theta/90)%4;
Mat3 matrix = NIL;
switch(choice){
    case 0: matrix = {{1.0,0.0,0.0}, {0.0,1.0,0.0}, {0.0,0.0,1.0}}; break;
    case 1: matrix = {{0.0,-1.0,0.0}, {1.0,0.0,0.0}, {0.0,0.0,1.0}}; break;
    case 2: matrix = {{-1.0,0.0,0.0}, {0.0,-1.0,0.0}, {0.0,0.0,1.0}}; break;
    case 3: matrix = {{0.0,1.0,0.0}, {-1.0,0.0,0.0}, {0.0,0.0,1.0}}; break;
    default: matrix = {{1.0,0.0,0.0}, {0.0,1.0,0.0}, {0.0,0.0,1.0}}; break;
}
  matMult(p, matrix);
  }

APPROACH＃3

   void rotate(Point *p, float theta){

int choice = ((int)theta/90)%4;
switch(choice){
    case 0: Mat3 matrix = {{1.0,0.0,0.0}, {0.0,1.0,0.0}, {0.0,0.0,1.0}}; break;
    case 1: Mat3 matrix = {{0.0,-1.0,0.0}, {1.0,0.0,0.0}, {0.0,0.0,1.0}}; break;
    case 2: Mat3 matrix = {{-1.0,0.0,0.0}, {0.0,-1.0,0.0}, {0.0,0.0,1.0}}; break;
    case 3: Mat3 matrix = {{0.0,1.0,0.0}, {-1.0,0.0,0.0}, {0.0,0.0,1.0}}; break;
    default: Mat3 matrix = {{1.0,0.0,0.0}, {0.0,1.0,0.0}, {0.0,0.0,1.0}}; break;
}
  matMult(p, matrix);
  }

他们都没有，工作和我得到大约50个关于}或}的错误;在错误的地方，我很确定不是问题。有任何想法吗？感谢

Answer 1

无法在C中分配矩阵，只能初始化矩阵或分配其各个组件。

有几种方法可以在这里进行，最简单的方法也许不是switch而是Mat3

的数组

static Mat3 const matrix[] = {
 {{1.0,0.0,0.0}, {0.0,1.0,0.0}, {0.0,0.0,1.0}},
 {{0.0,-1.0,0.0}, {1.0,0.0,0.0}, {0.0,0.0,1.0}},
 {{-1.0,0.0,0.0}, {0.0,-1.0,0.0}, {0.0,0.0,1.0}},
 {{0.0,1.0,0.0}, {-1.0,0.0,0.0}, {0.0,0.0,1.0}},
};

unsigned choice = ((unsigned)(theta/90))%4u;
matMult(p, matrix[choice]);

我添加了const，因为它看起来是你的第二个参数功能是只读的。
static只在真实之前分配并初始化了一次执行代码。

unsigned

choice保证%的算术确实会产生从0到3的值，而不会产生负值。

Answer 2

复合文字可以在C99之后使用。你可以写如下：，

void rotate(Point *p, float theta){
    int choice = ((int)theta/90)%4;

    switch(choice){
        case 0:  matMult(p, (Mat3){{1.0,0.0,0.0},  {0.0,1.0,0.0},  {0.0,0.0,1.0}}); break;
        case 1:  matMult(p, (Mat3){{0.0,-1.0,0.0}, {1.0,0.0,0.0},  {0.0,0.0,1.0}}); break;
        case 2:  matMult(p, (Mat3){{-1.0,0.0,0.0}, {0.0,-1.0,0.0}, {0.0,0.0,1.0}}); break;
        case 3:  matMult(p, (Mat3){{0.0,1.0,0.0},  {-1.0,0.0,0.0}, {0.0,0.0,1.0}}); break;
        default: matMult(p, (Mat3){{1.0,0.0,0.0},  {0.0,1.0,0.0},  {0.0,0.0,1.0}}); break;
    }
}

在条件或开关情况下，在C中初始化矩阵（2D阵列）

2 个答案: