The interviewer with ByteDance threw me an interview question and said that if I could answer it, other companies would have an 80% chance of passing the technical level

Hello everyone , I am a Java Big factory interview you .
Yes , No mistake. , Years ago, I went to byte for an interview ！

Let's share an interview episode

interviewer ：“ Are you married? ？”
I ：“ No, ”
interviewer ：“ I don't have ”

Get down to business , I won't talk about the specific resume writing and interview process , A lot of people share .
After the meeting, ask me when I can be on duty , I said to think about it again .

This interview can be described as from java Basic grammar 、JVM、 Multithreading to MySQL、spring、springboot、springcloud, There's more dubbo、mybatis、redis、 The bottom of the network IO、Linux、MQ、zookeeper、netty、 big data 、 Data structure and algorithm 、 And then to design patterns and so on .

After all, we are technical , What questions did you ask during the interview .

Question 1 ：CMS、G1 Three in the garbage collector ⾊ Mark do you understand ？

3、 ... and ⾊ The notation is ⼀ A waste recycling method , It can make JVM No hair ⽣ Or just for a short time ⽣ STW(Stop TheWorld), from ⽽ reach
To clear JVM Memory garbage ⽬ Of .
3、 ... and ⾊ The notation marks the face of the object ⾊ Divided into ⿊、 ash 、⽩, Three colors ⾊.
⿊⾊： The object has been marked , And all the attributes under the object have been marked ;
ash ⾊： The object has been scanned by the garbage collector , But there are still references in the object that have not been scanned ⽤ ;
⽩⾊： Indicates that the object has not been accessed by the garbage collector , I.e. not reachable ;

3、 ... and ⾊ Missing mark problem of marking

Question two ： If CPU send ⽤ The rate suddenly soared , How would you check ？

1、 Through the first top Command found consumption cpu very ⾼ The process of id( The assumption is 2732)
top The order is that we are Linux Most often ⽤ Of the orders of ⼀, It can display the execution in real time ⾏ Process CPU send ⽤ rate 、 Memory makes ⽤ Rate and system load . The upper part shows the statistical information of the system , The lower part shows the process of making ⽤ Rate statistics .

2、 Of board ⾏top -p 2732 Monitor the process separately
3、 In the 2 Step monitoring community ⾯ transport ⼊H, Get all thread information under the current process

4、 Find the consumption cpu special ⾼ Thread number （ The assumption is 2734）

5、 Of board ⾏jstack 2732 Make changes to the current process dump, Output all thread information
At the same time, the 4 The thread obtained in step ⼗ Hexadecimal number 2734 Turn into 16 Base number （AAE）, In the stack information ⾥⾯ Find the corresponding thread content

6、 Finally, read the thread information , Locate the specific code location

How are you seeing here ？

Question 3 ： Please outline AQS

yes ⽤ To build the basic framework for locks or other synchronous components ,⽐ Such as ReentrantLock、ReentrantReadWriteLock and
CountDownLatch Is based on AQS Realized . It makes ⽤ 了 ⼀ individual int The member variable represents the synchronization state , Through the built-in
FIFO Queue to complete the queue of the resource acquisition thread ⼯ do . It is CLH Queue lock ⼀ Two variants implement . It can achieve 2 Same species
Step ⽅ type ： exclusive , Shared .
AQS The main cause of ⽤⽅ Formula is inheritance ,⼦ Class by inheritance AQS And implement its abstraction ⽅ Method to manage synchronization status , Synchronizer's
Design is based on templates ⽅ Law mode , So if we want to achieve ⾃⼰ Synchronization of ⼯ Classes need to be overridden ⼏ A rewritable ⽅
Law , Such as tryAcquire、tryReleaseShared wait .
Designed like this ⽬ Is the synchronization component （⽐ Like a lock ） yes ⾯ To make ⽤ Of , It defines to make ⽤ The interface between the user and the synchronization component ⼝
（⽐ For example, two threads can be allowed to work together ⾏ visit ）, Hidden implementation details ; synchronizer ⾯ To the lock implementer , It simplifies
Lock implementation ⽅ type , The synchronization state management is blocked 、 Thread queuing 、 Wait and wake up and other underlying operations . This is a good way to isolate
Made ⽤ Areas of concern for developers and implementers .

In the internal ,AQS maintain ⼀ Shared resources state, Through the built-in FIFO To complete the queuing of the thread to obtain resources ⼯ do . This queue
from ⼀ individual ⼀ One of the Node Node composition , Every Node Node maintenance ⼀ individual prev lead ⽤ and next lead ⽤, Point to respectively ⾃⼰ Before
Drive and successor nodes , constitute ⼀ Two end two-way linked list .
At the same time Condition Related waiting queues , The node type is also Node, constitute ⼀ A one-way list .

Question 4 ：InnoDB⼀ Tree B+ How much can a tree hold ⾏ data ？

The problem is actually ⾮ Often simple ： about 2 Ten million
When a computer stores data , Have the most ⼩ Storage unit , In a computer, disk storage is the most ⼩ A unit is a sector ( That's good ⽐ We today
Tianjin ⾏ present ⾦ The circulation of ⼩ The unit is ⼀⽑),⼀ It's a sector ⼤⼩ yes 512 byte ,⽽⽂ Piece system （ for example XFS/EXT4） The most
⼩ A unit is a block ,⼀ Block ⼤⼩ yes 4k,⽽ For our InnoDB Storage engines also have ⾃⼰ The most ⼩ Storage unit ——⻚
（Page）,⼀ individual ⻚ Of ⼤⼩ yes 16K.

Innodb All data for ⽂ Pieces of （ The suffix is ibd Of ⽂ Pieces of ）, His ⼤⼩ It's always 16384（16k） Integer multiple .

The data in the data table is stored in ⻚ Medium , therefore ⼀ individual ⻚ How much can be stored in ⾏ Data? ？ hypothesis ⼀⾏ Data ⼤⼩ yes 1k, that
Well ⼀ individual ⻚ It can store 16 ⾏ Data like this .
about B+ Trees ⽽⾔, Only leaves ⼦ Nodes store data ,⾮ leaf ⼦ The node stores only the index information and the following information ⼀ The pointer letter of the layer node
Rest .⼀ individual ⾮ leaf ⼦ How many pointers can a node store ？
In fact, this is also very good , Let's assume that the primary key ID As usual ⽤ Of bigint type ,⻓ Degree is 8 byte ,⽽ The pointer ⼤⼩ stay InnoDB Source code
Set to 6 byte , such ⼀ common 14 byte , We ⼀ individual ⻚ How many such units can be stored in , In fact, it means how many hands there are , namely
16384/14=1170 individual .
Then we can work out ⼀ Tree ⾼ Degree is 2 Of B+ Trees , There is ⼀ A root node and if ⼲ Leaves ⼦ Nodes can store 117016=18720 This way
Data record of .
According to the same principle, we can calculate ⼀ individual ⾼ Degree is 3 Of B+ Trees can store ： 11701170*16=21902400 A note like this
record .

Can you answer these questions ？

One 、ActiveMQ What to do if the server goes down ？

Two 、JDK Dynamic proxy sum CGLIB Dynamic proxy differences ？

3、 ... and 、Redis The most appropriate scene ？

Friends who can see here must be interested in interviewing big factories , Or those who are learning to prepare for a big noodle factory , The notes range from introductory to advanced to advanced and practical cases , It can be said to be perfect .

give the thumbs-up All of my little friends finally got their favorite big factory offer Oh ༻ི( ᵒ̶̷ᵕ ᵒ̷̶)༄୭*ˈ