10
String str="abc" 一定会进入到字符串常量池吗？
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String str="abc" 一定会进入到字符串常量池吗？
String str="abc" 一定会进入到字符串常量池吗？
如果一定会，为什么我对 str 加锁，锁并没有用。
6 条回复 • 2021-07-08 20:25:59 +08:00
Yimkong 12 小时 17 分钟前 via iPhone
锁是加给对象的，你这个每次加锁都是都会锁一个新的 String 对象。
String str=new String("abc"),锁这个就行
String str=new String("abc"),锁这个就行
Cy1 12 小时 11 分钟前 
1
你是怎么加锁的？
public static void main(String[] args) {
String lock1 = "abc";
String lock2 = "abc";
new Thread(() -> {
synchronized (lock1) {
System.out.println("t1");
try {
TimeUnit.SECONDS.sleep(5);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}, "t1").start();
new Thread(() -> {
synchronized (lock2) {
System.out.println("t2");
try {
TimeUnit.SECONDS.sleep(5);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}, "t2").start();
}
其中一个线程会被阻塞，明显就是锁的是同一个对象，所以会进 字符串常量池
public static void main(String[] args) {
String lock1 = "abc";
String lock2 = "abc";
new Thread(() -> {
synchronized (lock1) {
System.out.println("t1");
try {
TimeUnit.SECONDS.sleep(5);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}, "t1").start();
new Thread(() -> {
synchronized (lock2) {
System.out.println("t2");
try {
TimeUnit.SECONDS.sleep(5);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}, "t2").start();
}
其中一个线程会被阻塞，明显就是锁的是同一个对象，所以会进 字符串常量池
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