英文:
Can't pass object into function
问题
所以我正在尝试制作一个程序，用于提供树莓派和加速度计/陀螺仪之间的接口。我需要这个函数尽可能频繁地更新，并且我想设置一个定时器来运行它。
以下是我的代码：
void updateGyro(Gyro gyro) {gyro.update();}int main() {if (gpioInitialise() < 0)return -1;// 创建新的陀螺仪对象Gyro gyro;// 设置一个定时器，每10毫秒更新一次gpioSetTimerFuncEx(0, 10, updateGyro, &gyro);time_sleep(10);gpioTerminate();}
但这似乎不起作用，我得到了错误：
gyro.cpp: In function ‘int main()’:gyro.cpp:113:31: error: invalid conversion from ‘void (*)(Gyro)’ to ‘gpioTimerFuncEx_t’ {aka ‘void (*)(void*)’} [-fpermissive]113 | gpioSetTimerFuncEx(0, 10, updateGyro, &gyro);| ^~~~~~~~~~| || void (*)(Gyro)In file included from gyro.cpp:2:/usr/include/pigpio.h:3751:55: note: initializing argument 3 of ‘int gpioSetTimerFuncEx(unsigned int, unsigned int, gpioTimerFuncEx_t, void*)’3751 | unsigned timer, unsigned millis, gpioTimerFuncEx_t f, void *userdata);| ~~~~~~~~~~~~~~~~~~^
我尝试将gyro.update函数直接传递给函数，但也没有起作用。有人知道如何修复这个问题吗？
这是我想要使用的函数的文档：
英文:
So I'm trying to make a program to provide a interface between a raspberry pi and a accelerometer/gyroscope. I need this function to update as often as possible and I want to set a timer for when it will run.
And here is my code:
void updateGyro(Gyro gyro) {gyro.update();}int main() {if (gpioInitialise() < 0)return -1;// Creates new gyro objectGyro gyro;// Sets a timer to update every 10 millisecondsgpioSetTimerFuncEx(0, 10, updateGyro, &gyro);time_sleep(10);gpioTerminate();}
But this doesn't seem to work and I get the error:
gyro.cpp: In function ‘int main()’:gyro.cpp:113:31: error: invalid conversion from ‘void (*)(Gyro)’ to ‘gpioTimerFuncEx_t’ {aka ‘void (*)(void*)’} [-fpermissive]113 | gpioSetTimerFuncEx(0, 10, updateGyro, &gyro);| ^~~~~~~~~~| || void (*)(Gyro)In file included from gyro.cpp:2:/usr/include/pigpio.h:3751:55: note: initializing argument 3 of ‘int gpioSetTimerFuncEx(unsigned int, unsigned int, gpioTimerFuncEx_t, void*)’3751 | unsigned timer, unsigned millis, gpioTimerFuncEx_t f, void *userdata);| ~~~~~~~~~~~~~~~~~~^
I've tried passing in the gyro.update function in directly into the function but that also didn't work. Does anyone know how to fix this?
Here is the documentation for the function I want to use:
答案1
得分: 6
void*不是任何数量或类型参数的占位符。它确切地表示单个指向未知类型的指针。updateGyro 不接受类型为 void* 的参数，它接受类型为 Gyro 的参数。
您需要更改您的函数以接受 void*，如下所示：
void updateGyro(void* arg) {Gyro* gyro = static_cast<Gyro*>(arg);gyro->update();}
英文:
void* is not a placeholder for any number of any arguments. It means exactly that, single pointer to unknown type. updateGyro doesn't accept argument of type void*, it accepts argument of type Gyro.
You need to change your function to accept void* instead:
void updateGyro(void* arg) {Gyro* gyro = static_cast<Gyro*>(arg);gyro->update();}
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