c++ - C++ 中的友元方法 "not declared in this scope"

Question

首先提供一些上下文，这是针对涉及信号量的赋值。我们要找到哲学家就餐问题的代码，让它运行起来，然后进行一些分析和操作。但是，我遇到了一个错误。

原始代码取自http://www.math-cs.gordon.edu/courses/cs322/projects/p2/dp/使用 C++ 解决方案。

我在 Code::Blocks 中收到的错误是

philosopher.cpp|206|error: 'Philosopher_run' was not declared in this scope|

此错误发生在行中:

if ( pthread_create( &_id, NULL, (void *(*)(void *)) &Philosopher_run,
         this ) != 0 )

我已查找过 pthread_create 方法，但无法修复此错误。如果有人能向我解释如何更正此错误，以及为什么会出现此错误，我将不胜感激。我已尝试仅提供相关代码。

class Philosopher
{
private:
    pthread_t   _id;
    int     _number;
    int     _timeToLive;

public:
    Philosopher( void ) { _number = -1; _timeToLive = 0; };
    Philosopher( int n, int t ) { _number = n; _timeToLive = t; };
   ~Philosopher( void )     {};
    void getChopsticks( void );
    void releaseChopsticks( void );
    void start( void );
    void wait( void );
    friend void Philosopher_run( Philosopher* p );
};

void Philosopher::start( void )
// Start the thread representing the philosopher
{
    if ( _number < 0 )
    {
    cerr << "Philosopher::start(): Philosopher not initialized\n";
    exit( 1 );
    }
    if ( pthread_create( &_id, NULL, (void *(*)(void *)) &Philosopher_run,
         this ) != 0 )
    {
    cerr << "could not create thread for philosopher\n";
    exit( 1 );
    }
};

void Philosopher_run( Philosopher* philosopher )

Answer 1

friend 声明不会使 friend 的名字在没有参数相关查找的情况下可见。

§7.3.1.2 [namespace.memdef] p3

[...] If a friend declaration in a nonlocal class first declares a class or function the friend class or function is a member of the innermost enclosing namespace. The name of the friend is not found by unqualified lookup or by qualified lookup until a matching declaration is provided in that namespace scope (either before or after the class definition granting friendship). [...]

意味着你应该把 void Philosopher_run( Philosopher* p ); 放在类之前(连同 Philosopher 的前向声明)，或者在类之后(while将友元声明保留在类中)。