450 Shenxin Mianjing 1

1、 When are copy constructors and assignment functions used , What should I pay attention to ？

• copy constructor ： Initialize a with an existing object New creation When an object is called .
• Assignment constructors ： Use an existing object against another There is already an object When assigning values, call .

Be careful ：

• When a pointer in a class points to heap space , These two constructors need to be noted that “ Shallow copy ” The problem of （ Two pointers point to the same heap space , The space is released twice ）.

2、extern "c" What's the difference between adding and not adding ？

function and Variable front Add extern Keywords difference ：

• extern External keywords ,extern You can modify variables or functions （ Function is defaulted by extern modification ）;
• Indicates that the definition of this variable or function is in other files , Prompt the compiler to find its definition in other files when encountering this variable and function .

Dynamic link library DLL Library plus extern “C” The role of ：

• stay C/C++ Mixed programming is used , stay DLL Add extern “C” Indicates that this function uses C Language .

3、 Memory alignment principles

What is memory alignment ？

• In many CPU Under the architecture ,CPU Instructions require that the starting address of the operation memory can be divided by the size of the operation memory ; Memory access that meets this requirement is called access aligned memory ;
• Generally speaking, the compiler will automatically align the memory .
• Memory alignment is the arrangement rule of data storage in memory , Easy to read by hardware .

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Memory alignment factor ：

#pragma pack(n) To set variables to n Byte alignment .

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Memory alignment rules ：

1、 The first data member of the structure is placed at the offset offset by 0 The place of , In the future, the alignment of each data member is in accordance with #pragma pack The specified value and The length of the data member itself in , Than Align smaller （ Offset to first address , The specified value needs to be a multiple of this small number ）.

2、 After the data members are aligned , class ( Structure or union ) Alignment itself , The alignment will follow #pragma pack The specified value and structure ( Or in association with ) Maximum data member length in （ The smaller one is aligned , The specified value needs to be a multiple of this small number .）

3、 Byte alignment is Determined at compile time , Once the decision is made, it will not change , Never change the size during operation .

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Why memory alignment ？

• The platform has good portability ;
• CPU High execution efficiency ;

example 1：

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example 2：

With 32 For example, bit system ;

Code 1：

struct A{
    
	short a;
	short b;
	short c;
};
 
struct B{
    
	struct A a;
	char c;
	int d;
};
 
printf("%d\n",sizeof(struct B));//-- Output  12

Code 2：

struct A{
    
	short a;
	short b;
	short c;
    char d;
};
 
struct B{
    
	struct A a;
	char c;
	int d;
};
 
printf("%d\n",sizeof(struct B));//-- Output  16

For code 1：

• struct A It was based on short 2 Byte Alignment ,struct A Itself is 6 Bytes , But put him in struct B In the middle of the day , Is due to struct B Is in accordance with the int 4 Byte alignment , So we need to struct B to struct A Separate out 2 Bytes of space is used to gather 4 Byte alignment , this 2 A byte is composed of struct B Provided , therefore struct B Members in can occupy , That is to say struct B In the char It can occupy , So the total is 12 Bytes .

For code 2：

• struct A according to short 2 Byte alignment ,struct A The size of itself is 8 byte , in other words struct A The last one char In the back 1 Bytes of space , But notice , this 1 The space of bytes is due to struct A Self aligned , It belongs to struct A Of , So will struct A Put in struct B In the middle of the day , although struct B There is one of them. char, But it can't be occupied struct A The remaining one in 1 Bytes of space , Only one more can be allocated 4 Bytes are used to store , After storage, there is 3 Bytes are not enough to store the following int 了 , redistribution 4 Bytes are used to store int, So the total is 16 Bytes .

4、memcpy Can the parameter be a structure ？

Sure ！

• because mem The first ones are memory copies , You can copy any data .（ Copy structure to array , Copy structure to structure , The array can be copied to the structure ）.

6、 A chessboard , There is a horse , Give a starting position and an ending position , How to find the shortest path ？

backtracking （ Like a sword finger offer–13. The range of motion of the robot ）, Only the number of shortest paths can be found .

7、epoll Characteristics and differences between horizontal trigger and edge trigger ？

Level trigger （ The default mode ）：

• If the kernel does not respond socket The data sent corresponding to the event is read （ If the buffer is too small ）, The next time epoll Will still put this socket The corresponding event is added to the response queue , Until the data in the buffer is read .

Edge trigger （ High speed mode ）：

• The kernel will not respond socket The data sent by the event is read , The next time epoll I won't prompt this again socket event , Until the next time there is data written, it will respond again .

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Combined with the following 8 The blocking and non blocking of can be seen ：

• Whether blocking or non blocking , When reading data, you need to use while Cycle to read all the data
• （LT The mode can be omitted while But it didn't work , Because the complete data cannot be processed at one time , You need to wait for the next response before continuing to receive the rest of the data , So it's still used once while Receive the data ）（ Because every time recv Receiving is not guaranteed to make the set length , It may be smaller than this length ）.

epoll Of ET（ Edge trigger ） Mode must use non blocking IO：

Because if blocking is used IO：

• Case one , Don't use while Read all the data , Then the data read after the next event may be the content left in the buffer of the previous event , This will affect the handling of this incident .
• The second case , Use while Read out all the data of each response event , that The last time recv Will be stuck here when , Because it's clogged , Then this thread will be invalidated , You can only wait for the client to send the next data read before responding , That is, this thread can only serve this client .

Just use while To ensure that the data is read , You can't use blocking IO.

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If you use it every time while Process the data together after reading it completely , There is no difference between the two modes ;

In the following cases ：

• If there is a file descriptor of interest but this event is not handled , that LT In mode epoll Will always return the response of this event , and ET In mode, it will not .

8、 Blocking and non blocking fd Of recv perhaps send The difference between ？

In blocking mode ：

• send： wait for send The function copies all data into the send buffer before returning .
• recv： If No data, just waiting , When there's data coming ,Recv Function copies the contents of the receive buffer to the application layer buffer in （ Be sure to pay attention to , Not all copies are returned here , Instead, as long as there is copied data, it will be returned , That is, it may be smaller than the length of data you want to accept ）.

In non blocking mode ：

• send： The function immediately returns , Between call and return Write data into the send buffer as much as possible . When to send is decided by the system .
• recv： If No data is returned directly , If there's data Copy the contents of the receive buffer to the application layer buffer in Just go back to ！（ It may not be all finished , Same as blocking ）.

PS： Although blocking and non blocking Recv The working principle of receiving data is similar , But the main difference is whether to return directly when there is no data to receive .

9、udp How to quickly locate clients that are not responding ？

• Send a broadcast , The client receives a reply ;
• because UDP Packets may be lost , You can send multiple broadcasts , List the clients that do not reply once as invalid clients .

10、 How to be in 40 One hundred million , Data is out of order , Give a number , How to find this number quickly ？

problem ：

• On a given set 4G Of PC Implemented on the machine , A contain 40 Hundreds of millions of unrepeated and unordered unsigned int Integers , Give an integer , Find a given number m, Is it in the file 40 Out of 100 million data .

Demand analysis ：

• int Type in the C++ Storage occupation in 4 individual Byte,32Bit;

• If defined in memory 40 One hundred million int Type array to read files , Occupation size :(40* 100000000*4/1024/1024/1024)G=14.901G.

• This is far beyond the memory limit of the machine , At this time, the conventional thinking is , Save data on disk , Read data into memory in batches , But this will Generate disks IO, For us who pursue speed , In this case, we don't consider .

For data , Is there a certain data or not , We can use 0 and 1 To express , This just matches binary 0 1 attribute ;

We know C++ in 1 int = 4 Byte = 32 Bit; In this case , The memory space we need is ,（40*100000000/8/1024/1024）M = 476.8！ It actually meets our requirements .

Specific ideas ：

1 individual int Occupy 4 Byte is 4*8=32 position , Then we only need to apply for one int The array length is int tmp[1+N/32] You can store all this data （ among N Represents the total number of searches to be made ,tmp Each element in is in memory 32 Bits can correspond to decimal numbers 0~31）

The following is bitmap algorithm ;（ Reference resources https://blog.csdn.net/Edward_LF/article/details/124614606）