小编Luk*_*uky的帖子

从在UI线程中运行代码的角度来看,之间有什么区别:

MainActivity.this.runOnUiThread(new Runnable() {
    public void run() {
        Log.d("UI thread", "I am the UI thread");
    }
});

要么

MainActivity.this.myView.post(new Runnable() {
    public void run() {
        Log.d("UI thread", "I am the UI thread");
    }
});

和

private class BackgroundTask extends AsyncTask<String, Void, Bitmap> {
    protected void onPostExecute(Bitmap result) {
        Log.d("UI thread", "I am the UI thread");
    }
}

我有一个单身人士课程:

public class Singleton {
    private static Singleton istance = null;

    private Singleton() {}

    public synchronized static Singleton getSingleton() {
        if (istance == null)
            istance = new Singleton();
        return istance;
    }

    public void work(){
            for(int i=0; i<10000; i++){
                Log.d("-----------", ""+i);
            }
    }
}

多个线程正在调用work()函数:

public class Main {

public static void main(String[] args) {

    new Thread (new Runnable(){
        public void run(){
            Singleton s = Singleton.getSingleton();
            s.work();}
    }).start();

    System.out.println("main thread");

    new Thread(new Runnable() { 
         public void run() {
             Singleton s = …

我注意到以下C代码给出了"警告:初始化从指针目标类型中丢弃限定符",但它仍然按预期编译和运行(输出'W'字符).

#include <stdio.h>
int main(int argc, char *argv[])
{
    char buffer[20] = {'H','e','l','l','o',' ','W','o','r','l','d','!','\0'};

    const char* p = &buffer[0];

    char* c = (p + 6);

    printf("%c\n",*c);
}

在C++中,相当类似的代码根本不会编译抱怨"错误:从'const char*'到'char*'的无效转换"

#include <iostream>
using namespace std;
int main()
{
   char buffer[20] = {'H','e','l','l','o',' ','W','o','r','l','d','!','\0'};

   const char* p = &buffer[0];

   char* c = p + 6;

   cout << *c;
   cout << endl;
   return 0;
}

是什么原因？

是否有可能修复C++代码以使其编译(和行为)就像它的C对应物一样？

更好的解释:感谢您的所有答案,但大多数人没有得到我真正的问题所以我会尝试更详细地解释.

我正在使用用C编写的库.标题中的函数原型是这样的:

void parse (const char* p, uint16_t len, uint8_t is_eof);

在这个函数的实现中,碰巧运行代码就好了

char* c = p …