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Accepted
Alexander Savushkin
Asked:2022-07-29 16:08:08 +0000 UTC2022-07-29 16:08:08 +0000 UTC

如何将带参数的函数传递给另一个函数？

  • 772
gfx.renderFF(message.info("Hello!", "I am console,", "how are you?"));

轮到时

void renderFF( void (*draw_fn)(String &a, String &b, String &c) )
{
  uint32_t time;
  time = millis() + 10;

  do {
    u8g2.clearBuffer();
    draw_fn();
    u8g2.sendBuffer();
  } while ( millis() < time );
}
c++ функции

2 个回答

  1. Best Answer

    C++11 风格解决方案：

    void renderFF( std::function<void(void)> draw_fn )
{ ...
   draw_fn();
 ...
}

gfx.renderFF( [&message](){ 
        message.info("Hello!", "I am console,", "how are you?");
    } );

    C风格：

    void renderFF( void (*draw_fn)(void*), void* draw_fn_args  )
{ ...
   draw_fn(draw_fn_args);
 ...
}

void renderFF_callback(void*arg)
{
    Message& message=*static_cast<Message*>(arg);
    message.info("Hello!", "I am console,", "how are you?");
};

gfx.renderFF( renderFF_callback, &message );
    • 0

  2. 您传递一个函数对象并调用它。该对象的参数可以设置在第一级（在调用renderFF之前的main中）、第二级、renderFF函数内部，以及第三级已经在 lambda 调用的代码中。

    [arg1]- 我们从主
    fun(arg2)函数中捕获参数- 我们将参数传递给renderFF
    info(arg1,arg2, "how are you?" )函数- 参数已经在 lambda 代码中

    # include <iostream>
# include <functional>

void renderFF( std::function<void(char const * arg2)> fun )
{
char const * arg2 = "I am console," ;
fun(arg2);
}

void info(char const*a,char const*b,char const*c){
  std::cout<<a<<b<<c<<std::endl;
}

int main()
{
  char const * arg1 = "Hello!" ;
  renderFF ( [arg1](char const * arg2) {
    info(arg1, arg2, "how are you?"); } );
}
    • 0

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