Question

我在做什么错了？

在第一个打印中，值很模糊，在第二个打印中都是零。

第一次打印-q：0.965926 0.000000 0.258819 0.000000

第二次打印-q：0.000000 0.000000 0.000000 0.000000

float *AnglesToQaternion(float roll, float pitch, float yaw) { // Convert Euler angles in degrees to quaternions

    static float q[4];

    roll  = roll  * DEG_TO_RAD;
    pitch = pitch * DEG_TO_RAD;
    yaw   = yaw   * DEG_TO_RAD;

    float t0 = cosf(yaw * 0.5);
    float t1 = sinf(yaw * 0.5);
    float t2 = cosf(roll * 0.5);
    float t3 = sinf(roll * 0.5);
    float t4 = cosf(pitch * 0.5);
    float t5 = sinf(pitch * 0.5);

    q[0] = t0 * t2 * t4 + t1 * t3 * t5;
    q[1] = t0 * t3 * t4 - t1 * t2 * t5;
    q[2] = t0 * t2 * t5 + t1 * t3 * t4;
    q[3] = t1 * t2 * t4 - t0 * t3 * t5;

    printf(">> q-1: %f %f %f %f \n", q[0], q[1], q[2], q[3]);

    return q;
}

float q[4] = *AnglesToQaternion(0, 30.0, 0); 
printf(">> q-2: %f %f %f %f \n", q[0], q[1], q[2], q[3]);

Answer 1

根本不返回数组。

using degree = float;

struct Angles {
    degree roll;
    degree pitch;
    degree yaw;
};

struct Quaternion {
    float i;
    float j;
    float k;
    float l;
};

Quaternion angles_to_quaternion(Angles angles) 
{ 
    float yaw = angles.yaw * DEG_TO_RAD;
    float pitch = angles.pitch * DEG_TO_RAD;
    float roll = angles.roll * DEG_TO_RAD;

    float t0 = cosf(yaw * 0.5);
    float t1 = sinf(yaw * 0.5);
    float t2 = cosf(roll * 0.5);
    float t3 = sinf(roll * 0.5);
    float t4 = cosf(pitch * 0.5);
    float t5 = sinf(pitch * 0.5);

    return {
        t0 * t2 * t4 + t1 * t3 * t5,
        t0 * t3 * t4 - t1 * t2 * t5,
        t0 * t2 * t5 + t1 * t3 * t4,
        t1 * t2 * t4 - t0 * t3 * t5,
    };
}

Answer 2

在代码更改最少的情况下，函数和q应该如下所示：

#include <array>

std::array<float, 4> AnglesToQaternion(float roll, float pitch, float yaw)
{
    std::array<float, 4> q{ 0,0,0,0 };

＆使用如下功能：

std::array<float, 4> q{ 0,0,0,0 };

q = AnglesToQaternion(0, 30.0, 0);

Answer 3

解决方案1：如果要能够从外部访问数组，则必须在函数内部动态创建数组。然后，您必须考虑删除它，以避免内存泄漏。

float *AnglesToQaternion(float roll, float pitch, float yaw) { // Convert Euler angles in degrees to quaternions

    float *q = new float[4];
    // Or C-style: float *q = malloc(sizeof(float)*4);

    roll  = roll  * DEG_TO_RAD;
    pitch = pitch * DEG_TO_RAD;
    yaw   = yaw   * DEG_TO_RAD;

    float t0 = cosf(yaw * 0.5);
    float t1 = sinf(yaw * 0.5);
    float t2 = cosf(roll * 0.5);
    float t3 = sinf(roll * 0.5);
    float t4 = cosf(pitch * 0.5);
    float t5 = sinf(pitch * 0.5);

    q[0] = t0 * t2 * t4 + t1 * t3 * t5;
    q[1] = t0 * t3 * t4 - t1 * t2 * t5;
    q[2] = t0 * t2 * t5 + t1 * t3 * t4;
    q[3] = t1 * t2 * t4 - t0 * t3 * t5;

    printf(">> q-1: %f %f %f %f \n", q[0], q[1], q[2], q[3]);

    return q;
}

int main() {
    float *q = AnglesToQaternion(0, 30.0, 0); 
    printf(">> q-2: %f %f %f %f \n", q[0], q[1], q[2], q[3]);
    delete[] q;
    // Or C-style: free(q);
}

解决方案2：您可以创建静态数组，然后将其第一个元素的地址传递给函数。

void AnglesToQaternion(float *q, float roll, float pitch, float yaw) { // Convert Euler angles in degrees to quaternions

    roll  = roll  * DEG_TO_RAD;
    pitch = pitch * DEG_TO_RAD;
    yaw   = yaw   * DEG_TO_RAD;

    float t0 = cosf(yaw * 0.5);
    float t1 = sinf(yaw * 0.5);
    float t2 = cosf(roll * 0.5);
    float t3 = sinf(roll * 0.5);
    float t4 = cosf(pitch * 0.5);
    float t5 = sinf(pitch * 0.5);

    q[0] = t0 * t2 * t4 + t1 * t3 * t5;
    q[1] = t0 * t3 * t4 - t1 * t2 * t5;
    q[2] = t0 * t2 * t5 + t1 * t3 * t4;
    q[3] = t1 * t2 * t4 - t0 * t3 * t5;

    printf(">> q-1: %f %f %f %f \n", q[0], q[1], q[2], q[3]);
}

int main() {
    float q[4];
    AnglesToQaternion(&q[0], 0, 30.0, 0); 
    printf(">> q-2: %f %f %f %f \n", q[0], q[1], q[2], q[3]);
}

第二种解决方案在性能上要好得多（如果可能，应尽可能避免使用new / delete）。

Answer 4

正如您在问题下的注释中提到的那样，您不能真正返回数组，而是返回指针。如果您确实希望将其用作返回值，请确保在使用该函数之前为其分配内存。像这样：

float *f_array = malloc(sizeof(float)*lengthArray);

另一种方法是将数组作为参数传递给函数，然后将其填充到函数中。

void AnglesToQaternion(float roll, float pitch, float yaw, float resultArray[])

Answer 5

将main（）中本地q的地址传递给AnglesToQaternion以便填充。

void AnglesToQaternion（浮点滚动，浮点俯仰，浮点偏航，浮点*四元数）

Answer 6

Caleth已经写出了答案的答案。但是我认为您需要开始考虑使用C ++而不是C。这是一个完整的示例：

Android.Support.V4.App.Fragment fragment = null;
fragment = Fragment3.NewInstance();

 if (fragment == null)
            return;

       SupportFragmentManager.BeginTransaction()
          .Replace(Resource.Id.content_frame, fragment)
          .Commit();

从函数c ++

6 个答案: