小编roy*_*las的帖子

我想知道是否可以在结构内实现通用引用。代码是：

struct Foo1
  ia :: Int
end


struct Foo2{UNKNOWNTYPE}
  ref :: UNKNOWNTYPE
  ib :: Int
  
  function Foo2{UNKNOWNTYPE}(ref::UNKNOWNTYPE,ib::Int)
    o = new{UNKNOWNTYPE}(ref,ib) 
    return o
  end
end

foo1 = Foo1();
foo2 = Foo2{Foo1}(foo1,1)

在上面的代码中，struct Foo2 中变量的类型ref直到运行时才确定。上面的代码不起作用，它显示：“LoadError("main.jl", 6, UndefVarError(:UNKNOWNTYPE))”。

我只是学习函数指针,并想测试它如何与成员函数一起工作.以下代码的编译在标记的位置失败.

# include <iostream>
# include <stdio.h>
using namespace std ;

class TMyClass {
public:
  int DoIt ( float a, char b, char c ) {
    cout << " TMyClass::DoIt " << endl ;
    return a + b + c ;
  }

  int DoMore ( float a, char b, char c ) {
    cout << " TMyClass::DoMore " << endl ;
    return a - b + c ;
  }

  int ( TMyClass::*pt2Member ) ( float, char, char ) ;

  int …

以下代码中的前两行输出是两个空白行,第三行和第四行是两个不相等的大数字,如:19147336 19147192

class A {
public:
    A() : m_i(0) { }
protected:
    int m_i;
};

class B {
public:
    B() : m_d(0.0) { }
protected:
    double m_d;
};

int main() {
    A *pa = new A;
    B *pb = new B;
    std::cout << reinterpret_cast<char*>(pa) << std::endl;
    std::cout << reinterpret_cast<char*>(pb) << std::endl;
    std::cout << (int)reinterpret_cast<char*>(pa) << std::endl;
    std::cout << (int)reinterpret_cast<char*>(pb) << std::endl;

    return 0;
}

我想知道上面代码中reinterpret_cast的返回是什么.谢谢!

以下是我的代码的主要部分(采用"高级linux编程",清单3.7):

void clean_up_child_process ( int signal_number ) {
  /* Clean up the child process. */
  int status ;
  wait ( &status ) ;

  printf ( " wait finished \n" ) ;

  /* Store its exit status in a global variable. */
  child_exit_status = status ;
}

int main() {
  /* Handle SIGCHLD by calling clean_up_child_process. */
  pid_t child_pid ;
  struct sigaction sigchld_action ;
  memset ( &sigchld_action, 0, sizeof(sigchld_action) ) ;
  sigchld_action.sa_handler = &clean_up_child_process ;

  sigaction ( SIGCHLD, &sigchld_action, NULL ) …