小编Luc*_*cas的帖子

我正在Linux中学习fork（），这是我的程序：

  1 #include <stdio.h>
  2 #include <stdlib.h>
  3 #include <unistd.h>
  4 int main(void){
  5         int pid;
  6         pid = fork();
  7         if(pid < 0){
  8                 exit(1);
  9         }
 10         if(pid == 0){
 11                 fork();
 12                 fork();
 13                 printf("pid:%d ppid:%d\n",getpid(),getppid());
 14                 exit(0);
 15                 }
 16         else{
 17                 printf("parent pid:%d ppid:%d\n",getpid(),getppid());
 18                 exit(0);
 19         }       
 20        
 21 }

有时效果很好，结果如下：

./test1.out
parent pid:27596 ppid:21425
pid:27599 ppid:27597
pid:27597 ppid:27596
pid:27598 ppid:27597
pid:27600 ppid:27598

但是结果并不一致，通常是这样的：

parent pid:27566 ppid:21425
pid:27567 ppid:27566
pid:27568 ppid:27567
pid:27569 ppid:1599
pid:27570 …

这个问题已经折磨了我两天了，如果没有python3-dev我就无法安装任何模块，有人可以告诉我如何解决这个问题吗？

The following packages have unmet dependencies:
     python3-dev : Depends: python3 (= 3.5.1-3) but 3.5.1-4 is to be installed
                   Depends: libpython3-dev (= 3.5.1-3) but it is not going to be installed
                   Depends: python3.5-dev (>= 3.5.1-2~) but it is not going to be installed
    E: Unable to correct problems, you have held broken packages.

class Glyph {
    void draw() {
        print("Glyph.draw()");
    }

    Glyph() {
        print("Glyph() before draw()");
        draw();
        print("Glyph() after draw()");
    }

}
class RoundGlyph extends Glyph{
    private int radius = 1;
    RoundGlyph(int r){
        radius = r;
        print("RoundGLyph.draw(), radius = " + radius);
    }
    void draw(){
        print("radius:" + radius);
    }
    public static void main(String[] args){
        new RoundGlyph(5);

    }

}

//Glyph() before draw()
radius:0
Glyph() after draw()
RoundGLyph.draw(), radius = 5

代码在上面.

由于draw()不是静态的,因此必须有一个隐含的参数(this).虽然在这种情况下,在Glyph的构造函数中调用draw(),所以我想知道这个"隐式参数"是什么. 据我所知,当我调用tf()时,t为T类型,编译器会把它变成Tf(t).

结果表明,在我看来,它是一个RoundGlyph作为这个参数提供.但这怎么可能？显然,调用draw()时不会创建RoundGlyph.