我正在用C ++为ESP8266编写一个小程序,遇到了麻烦。 我创建了一个Led类表单处理LED。想法是该类应处理眨眼函数。为此,我使用了一个名为Ticker的库。
Ticker中的一个函数,attach_ms需要回调,而我无法使它与非静态成员函数一起工作。
这是我的头文件:
#ifndef led_h
#define led_h
#include <Arduino.h>
#include <Ticker.h>
#include "debugutils.h"
#define tickLength 100
enum class LedState {
OFF,
ON,
SLOW_BLINK,
FAST_BLINK
};
class Led {
public:
Led(Ticker *tick, uint8_t ledPin, int slowBlinkTime, int fastBlinkTime);
void on();
void off();
void slowBlink( );
void fastBlink( );
private:
uint8_t pin;
int counter;
int slowNoBlinkTicks;
int fastNoBlinkTicks;
LedState state;
void ledOn();
void ledOff();
void ledInvert();
void clean();
void blink(int par);
void tickerCallbackLed();
};
#endif
这是我的代码文件:
#include "led.h"
void Led::ledOn() {
digitalWrite(pin, HIGH);
}
void Led::ledOff() {
digitalWrite(pin, LOW);
}
void Led::ledInvert() {
digitalWrite(pin, !digitalRead(pin));
}
void Led::clean() {
counter = 0;
}
void Led::blink(int par) {
if (counter > par) {
ledInvert();
counter = 0;
}
else {
counter++;
}
}
void Led::tickerCallbackLed() {
switch (state) {
case LedState::OFF : break;
case LedState::ON : break;
case LedState::SLOW_BLINK : blink(slowNoBlinkTicks); break;
case LedState::FAST_BLINK : blink (fastNoBlinkTicks); break;
default : break;
};
};
void Led::on() {
ledOn();
state = LedState::ON;
};
void Led::off(){
ledOff();
state = LedState::OFF;
};
void Led::slowBlink(){
clean();
ledInvert();
state = LedState::SLOW_BLINK;
};
void Led::fastBlink(){
clean();
ledInvert();
state = LedState::FAST_BLINK;
};
Led::Led(Ticker *tick, uint8_t ledPin, int slowBlinkTime, int fastBlinkTime) {
tick->attach_ms(tickLength, std::bind(&Led::tickerCallbackLed,this));
slowNoBlinkTicks = slowBlinkTime/tickLength;
fastNoBlinkTicks = fastBlinkTime/tickLength;
pinMode(ledPin,OUTPUT);
digitalWrite(ledPin,LOW);
pin = ledPin;
state = LedState::OFF;
counter = 0;
}
此行给出了编译错误,我不知道如何解决。尝试遵循我在互联网上找到的所有“建议”。
tick->attach_ms(tickLength, std::bind(&Led::tickerCallbackLed,this));
答案 0 :(得分:2)
根据this version of Ticker.h
,Ticker::attach_ms()
被重载以接受std::function<void(void)>
或void (*)(TArg)
作为回调:
typedef void (*callback_with_arg_t)(void*);
typedef std::function<void(void)> callback_function_t;
void attach_ms(uint32_t milliseconds, callback_function_t callback)
{
_callback_function = callback;
attach_ms(milliseconds, _static_callback, (void*)this);
}
template<typename TArg>
void attach_ms(uint32_t milliseconds, void (*callback)(TArg), TArg arg)
{
static_assert(sizeof(TArg) <= sizeof(uint32_t), "attach_ms() callback argument size must be <= 4 bytes");
uint32_t arg32 = (uint32_t)arg;
_attach_ms(milliseconds, true, reinterpret_cast<callback_with_arg_t>(callback), arg32);
}
在第一种情况下,您可以将lambda与std::function
一起使用,根本不需要std::bind()
:
tick->attach_ms(tickLength, [this](){ this->tickerCallbackLed(); });
在第二种情况下,callback
接受一个用户定义的参数,该参数传递给Ticker::attach_ms()
,因此您可以将this
作为该参数传递(如上所示) ,正是std::function
的{{1}}版本在内部所做的操作):
Ticker::attach_ms()
但是请注意,class Led {
...
private:
...
static void staticTickerCallbackLed(Led *pThis);
void tickerCallbackLed();
...
};
void Led::staticTickerCallbackLed(Led *pThis)
{
pThis->TickerCallbackLed();
}
...
tick->attach_ms(tickLength, &Led::staticTickerCallbackLed, this);
不允许使用大于4个字节的回调参数,这意味着在为指针为8个字节的64位进行编译时,这两种方法都不起作用!恕我直言,这似乎是内部Ticker::attach_ms()
方法实现中的错误,该方法将回调参数作为Ticker::_attach_ms()
而不是uint32_t
:
uintptr_t