C ++ pointer Essence

To understand pointers, there may be more or less complex types.

To fully understand a complex type, it is actually very easy to understand a complex type.

There will be many operators in the types, they are also like normal expressions, There is a priority

Level, the priority is the same as the operation priority, so I have summarized its principles:

Starting from the variable name, it is analyzed step by step based on the combination of operator priorities.

Next, let's start from a simple type and analyze it slowly:

Int P; // This is a common integer variable int * P; // starts from P and starts with *. Therefore, P is a // pointer, then combined with int, it indicates that the content type pointed to by the pointer is int type. therefore, P is a pointer to int P [3] That returns the integer/type data; // It starts from P and is first combined, P is a number // group, and then combined with int, it indicates that the elements in the array are integer, therefore, P is an array of integer data. int * P [3]; // It starts from P and combines with [] First, because it has a higher priority, therefore, P is an array and then combined with *, indicating that the elements in the // array are pointer types, and then combined with int, // indicates that the pointer points to an integer type, so // P is an array of INT (* P) [3] consisting of pointers that return integer data; // start with P and first combine with *, indicating that p is a pointer // and then combine with [] (and "()", which can be ignored, it only indicates that the pointer is directed to a // array and then combined with int, indicating that the elements in the array are // integer. therefore, P is a pointer int ** P to a number/group composed of integer data; // It starts from P and is first combined with *, indicating that p is a pointer, then, // is combined with *, indicating that the Pointer Points to the pointer, and then combined with int, indicating that the Pointer Points to the integer data. because second-level pointers and more advanced pointers are rarely used // in complex types, we will not consider multi-level pointers for more complex types. We will only consider first-level pointers at most. int P (INT); // starting from P, it is first combined with (), indicating that p is a function and then analyzed in, it indicates that the function has an integer Variable Parameter // and is then combined with an external int, indicating that the return value of the function is // an integer data int (* p) (INT ); // starting from P, it is first combined with the pointer, indicating that p is a pointer, and then combined with // (), indicating that the pointer points to a function, and then () the combination of '// int' indicates that the function has an int-type parameter, and is then combined with the outermost // int, indicating that the return type of the function is an integer, therefore, P refers to the int * (* P (INT) [3] pointer to a function with an integer parameter and an integer type returned. // you can skip this step first, without looking at this type, it is too complicated. // starting from P, it is first combined with (), indicating that p is a function, then it is included in // inbound () and combined with Int, it indicates that the function has an integer variable // parameter, and then it is combined with the external *, indicating that the function returns a // pointer, and then goes to the outermost layer, first in combination, description // The returned Pointer Points to an array, and then combines it with *. The elements in the // clear array are pointers, and then combine them with Int, description: The content pointed to by the needle is an integer data. therefore, P is a function that refers to a // integer and returns a pointer variable pointing to an array composed of integer pointer variables.

It is almost the same here, and we have so many tasks. We have understood these types and other

The type is also a small dish for us, but we generally do not use too complex types, that will

This greatly reduces the readability of the program. Please use it with caution. The above several types are enough for us to use.

1. Detailed pointer

A pointer is a special variable. The value stored in it is interpreted as an address in the memory. To understand a pointer, we need to understand the four aspects of the pointer: pointer type, pointer type, pointer value, memory area pointed to by pointer, memory area occupied by pointer itself. Let's explain it separately.

First, declare several pointers for example:

Example 1:

　　

(1)int*ptr; (2)char*ptr; (3)int**ptr; (4)int(*ptr)[3]; (5)int*(*ptr)[4]; 

1. pointer type

From the syntax perspective, you only need to remove the pointer name in the pointer declaration statement, and the rest is the pointer type. This is the type of the pointer. Let's take a look at the type of each pointer in Example 1:

(1) int * PTR; // the pointer type is int *

(2) char * PTR; // the pointer type is char *

(3) int ** PTR; // the pointer type is int **

(4) int (* PTR) [3]; // the pointer type is int (*) [3]

(5) int * (* PTR) [4]; // the pointer type is int * (*) [4]

How is it? Is it easy to find the pointer type?

2. Type pointed to by pointer

When you access the memory area pointed to by the pointer, the type pointed to by the pointer determines what the compiler will regard the content in the memory area.

In terms of syntax, you only need to remove the pointer name and the pointer declarative * on the left of the name in the pointer declaration statement, and the rest is the type pointed to by the pointer. For example:

(1) int * PTR; // The Pointer Points to an int type.

(2) char * PTR; // The Pointer Points to a char type.

(3) int ** PTR; // The type pointed to by the pointer is int *

(4) int (* PTR) [3]; // The type pointed to by the pointer is int () [3]

(5) int * (* PTR) [4]; // The type pointed to by the pointer is int * () [4]

In pointer arithmetic operations, the type pointed to by the pointer has a great effect.

The pointer type (the pointer type) and the pointer type are two concepts. When you get familiar with C, you will find that, the concept of "type", which is mixed with pointers, is divided into two concepts: "pointer type" and "pointer type", which are one of the key points of mastering pointers. I have read a lot of books and found that some poorly written books bring together the two concepts of pointers. Therefore, the books seem to have conflicts and become more confused.

3. pointer value ---- or the memory zone or address pointed to by the pointer

The pointer value is the value stored by the pointer itself. This value will be treated as an address by the compiler rather than a general value. In a 32-bit program, the value of all types of pointers is a 32-bit integer, because the 32-bit program's memory address is all 32-bit long. The memory area pointed to by the pointer starts from the memory address represented by the pointer value, and the length is a memory area of Si zeof (type pointed to by the pointer. Later, we will say that the value of a pointer is XX, which means that the pointer points to a memory area with xx as the first address. We will say that a pointer points to a memory area, it is equivalent to saying that the pointer value is the first address of the memory area.

The memory zone pointed to by the pointer and the type pointed to by the pointer are two completely different concepts. In example 1, the pointer points to a type that already exists, but since the pointer has not been initialized, the memory zone it points to does not exist, or it is meaningless.

In the future, every time you encounter a pointer, you should ask: What is the type of this pointer? What is the pointer type? Where does this pointer point? (Important)

4. Memory occupied by pointers

How much memory does the pointer occupy? You just need to use the sizeof function (pointer type) to test it. On a 32-bit platform, the pointer occupies 4 bytes.

The memory occupied by pointers can be used to determine whether a pointer expression (which will be explained later) is a left value.

2. arithmetic operations on pointers

The pointer can be added or subtracted from an integer. The meaning of this operation of pointer is different from that of addition and subtraction of common values, in units. For example:

Example 2:

Char A [20]; int * PTR = (int *) A; // forced type conversion does not change the type PTR ++ of;

In the preceding example, the pointer PTR is of the int type * and it points to the int type. It is initialized to the integer variable. In the next 3rd sentences, the pointer PTR is added with 1, and the compiler processes it like this: it adds the value of the pointer PTR with sizeof (INT), in a 32-bit program, 4 is added because int occupies 4 bytes in 32-bit programs. Because the address is in bytes, the address pointed by PTR is increased by four bytes from the address of the original variable A to the high address.

Because the length of the char type is one byte, the PTR originally points to the four bytes starting from Unit 0th of array, this point points to the four bytes starting from Unit 4th in array.

We can use a pointer and a loop to traverse an array. For example:

Example 3:

int array[20]={0}; int *ptr=array; for(i=0;i<20;i++) { 　(*ptr)++; 　ptr++； }

In this example, the value of each unit in the integer array is added to 1. Since every loop adds a pointer PTR to one unit, each loop can access the next unit of the array.

Let's look at the example:

Example 4:

char a[20]="You_are_a_girl";int *ptr=(int *)a;  ptr+=5;

In this example, PTR is added with 5, and the compiler processes it like this: add the value of the pointer PTR to the value of the 5-bysizeof (INT ), in the 32-bit program, 5 is multiplied by 4 = 20. Because the address unit is byte, the current PTR points to the address, compared to the address pointed by the PTR after the addition of 5, move 20 bytes to the high address. In this example, if the PTR before 5 is not added, it points to the four bytes starting with unit 0th of array A. After 5 is added, the PTR points out of the valid range of array. Although this situation may cause problems in applications, it is possible in terms of syntax. This also reflects the flexibility of pointers.

If in the above example, PTR is subtracted from 5, the processing process is similar, except that the PTR value is subtracted from 5 by sizeof (INT ), the new PTR points to an address that moves 20 bytes to the lower address direction than the original PTR.

Here is another example: (A misunderstanding)

Example 5:

#include<stdio.h>int main(){char a[20]=" You_are_a_girl";char *p=a;char **ptr=&p;//printf("p=%d\n",p);//printf("ptr=%d\n",ptr);//printf("*ptr=%d\n",*ptr);printf("**ptr=%c\n",**ptr);ptr++;//printf("ptr=%d\n",ptr);//printf("*ptr=%d\n",*ptr);printf("**ptr=%c\n",**ptr);}

Misunderstanding 1. Output answer: Y and O

Misunderstanding: PTR is a second-level pointer of char. When executing PTR ++;, it will add a sizeof (char) to the pointer, so the above result is output, this may be the result of a small number of people.

Misunderstanding 2. Output answer: Y and

Misunderstanding: PTR points to a char * type. When executing PTR ++;, it will add a sizeof (char *) to the pointer (someone may think this value is 1, then you will get the answer to misunderstanding 1. This value should be 4, refer to the previous content), that is, & P + 4; isn't a single value operation pointing to the fifth element in the array? Isn't the output result The Fifth Element in the array? The answer is no.

Positive solution: the PTR type is Char ** and the pointer type is char *. The pointer address is P (& P). When PTR ++ is executed, it will add a sizeof (char *) to the pointer, that is, & P + 4; Where does * (& P + 4) point to? Ask God, or he will tell you where it is? Therefore, the final output is a random value, which may be an invalid operation.

Summary:

After a pointer ptrold is added (subtracted) by an integer N, the result is a new pointer ptrnew. The type of ptrnew is the same as that of ptrold, ptrnew points to the same type as ptrold. The value of ptrnew will increase (decrease) n times of sizeof (type pointed to by ptrold) bytes than the value of ptrold. That is to say, ptrnew will point to a memory area that moves N x sizeof (the type indicated by ptrold) bytes to the high (low) Address direction than the memory area pointed to by ptrold.

Add or subtract pointer and pointer:

The two pointers cannot be used for addition operations. This operation is invalid because after addition, the result is directed to an unknown place and meaningless. The two pointers can be used for subtraction, but they must be of the same type. They are generally used in arrays.

3. Operators & and *

Here & is the address operator, and * is the indirect operator.

& The operation result of A is a pointer, And the pointer type is a plus *. The Pointer Points to the type, and the Pointer Points to the address, that is the address of.

* P's computation results are varied. In short, the result of * P is what P points to. It has these characteristics: its type is P-pointed, and its occupied address is the address pointed to by P.

Example 6:

Int A = 12; int B; int * P; int ** PTR; P = & A; // The result of & A is a pointer, whose type is int *, the type is // int, and the address is. * P = 24; // * P result. Here its type is int, and the address it occupies is the address pointed to by // P. Obviously, * P is variable. PTR = & P; // The result of & P is a pointer. the pointer type is p type plus *, // here is int **. The Pointer Points to the p type, which is int * In *. The Pointer Points to the address of P itself. * PTR = & B; // * PTR is a pointer, and the result of & B is also a pointer. The types of the two pointers // are the same as the types pointed, therefore, it is no problem to use & B to assign/value to * PTR. ** PTR = 34; // * the result of PTR is what PTR points to. Here it is a pointer. // perform another operation on this pointer, the result is an int type variable.

　　

4. pointer expression

If the result of an expression is a pointer, the expression is called a pointer table.

Below are some examples of pointer expressions:

Example 7:

Int A, B; int array [10]; int * pA; Pa = & A; // & A is a pointer expression. Int ** PTR = & PA; // & PA is also a pointer expression. * PTR = & B; // * PTR and & B are pointer expressions. Pa = array; PA ++; // This is also a pointer expression.

Example 8:

Char * arr [20]; char ** Parr = arr; // If arr is regarded as a pointer, arr is also a pointer expression char * STR; STR = * Parr; // * Parr is the pointer expression STR = * (Parr + 1); // * (Parr + 1) is the pointer expression STR = * (Parr + 2 ); // * (Parr + 2) is a pointer expression.

Since the result of a pointer expression is a pointer, the pointer expression also has four elements of the pointer: the pointer type, the type pointed to by the pointer, and the memory area pointed to by the pointer, memory occupied by the pointer itself.

Well, when the result pointer of a pointer expression has clearly occupied the memory of the pointer itself, the pointer expression is a left value, otherwise it is not a left value. In Example 7, & A is not a left value because it does not occupy clear memory. * PTR is a left value, because * PTR pointer occupies the memory. In fact, * PTR is the pointer Pa. Since PA already has its own location in the memory, * PTR certainly has its own position.

5. Relationship between arrays and pointers

The array name can be regarded as a pointer. See the following example:

Example 9:

Intarray [10] = {,}, value; value = array [0]; // It can also be written as: value = * array; value = array [3]; // It can also be written as: value = * (array + 3); value = array [4]; // It can also be written: value = * (array + 4 );

In the above example, the array name array represents the array itself, and the type is int [10]. However, if we regard array as a pointer, it points to the 0th units of the array, and the type is int *, the Type pointed to is the type of the array unit, that is, Int. Therefore, it is not surprising that * array is equal to 0. Similarly, array + 3 is a pointer to the array's 3rd units, so * (array + 3) is equal to 3. Others.

Example 10:

Char * STR [3] = {"Hello, thisasample! "," Hi, Goodmorning. "," helloworld "}; chars [80]; strcpy (S, STR [0]); // It can also be written as strcpy (S, * Str); strcpy (S, STR [1]); // It can also be written as strcpy (S, * (STR + 1); strcpy (S, STR [2]); // It can also be written as strcpy (S, * (STR + 2 ));

In the preceding example, STR is a three-unit array, and each unit of the array is a pointer, each pointing to a string. If the pointer array name STR is used as a pointer, it points to the cell 0th of the array. Its type is Char **, and it points to char *.

* STR is also a pointer. Its type is char *. It points to Char and its address is the string "Hello, thisasample! ", That is, the address of 'H. Note: The string is equivalent to an array, which is stored in the memory as an array, except that the string is an array constant and the content cannot be changed and can only be the right value. if it is regarded as a pointer, it is a constant pointer and a pointer constant.

STR + 1 is also a pointer pointing to unit 1st of the array. Its type is Char **, and its type is char *.

* (STR + 1) is also a pointer. Its type is char * and it points to Char, which points to the first character 'h' of "Hi, Goodmorning'

　　

The following is a summary of the array name (which is also stored in the array:

If an array type array [N] is declared, the array name array has two meanings: First, it represents the entire array, and its type is type [N]; second, it is a constant pointer. the pointer type is type *, and the Pointer Points to type, that is, the type of the array unit, the Pointer Points to the memory zone, which is the unit 0th of the array. the pointer occupies a separate memory zone. Note that it is different from the memory zone occupied by the array unit 0th. The value of this pointer cannot be modified, that is, the expression similar to array ++ is incorrect.

In different expressions, array names can play different roles.

In the expression sizeof (array), the array name array represents the array itself. Therefore, the sizeof function measures the size of the entire array.

In the expression * array, array acts as a pointer. Therefore, the result of this expression is the value of unit 0th of the array. Sizeof (* array) measures the size of the array unit.

Expression Array + N (where n = 0, 1, 2 ,.....) in, array is a pointer, so the result of array + N is a pointer, its type is type *, it points to type, it points to the array number n. Therefore, sizeof (array + n) measures the pointer size. In the 32-bit program, the result is 4.

Example 11:

Int array [10]; int (* PTR) [10]; PTR = & array;: in the preceding example, PTR is a pointer and its type is int (*) [10]. the Type pointed to is int [10]. We use the first address of the entire array to initialize it. In the statement PTR = & array, array represents the array itself. As mentioned in this section, the sizeof () function is used to determine whether the sizeof (pointer name) is the size of the pointer's own type or the size of the type pointed to by the pointer? The answer is the former. For example: int (* PTR) [10]; In a 32-bit program, there are: sizeof (INT (*) [10]) = 4 sizeof (INT [10]) = 40 sizeof (PTR) = 4

In fact, sizeof (object) measures the size of the object's own type, rather than the size of other types.

6. Relationship between pointer and Structure Type

You can declare a pointer to a structure object.

Example 12:

Struct mystruct {int A; int B; int C ;}; struct mystruct SS ={20, 30, 40}; // declares the structure object SS, initialize the SS members to 20, 30, and 40. Struct mystruct * PTR = & SS; // declares a pointer to the structure object ss. Its type is // mystruct *, and it points to the type mystruct. Int * pstr = (int *) & SS; // declares a pointer to the structure object ss. However, pstr and // The type PTR to which it is directed are different.

　　

How can I access the three member variables of SS through the pointer PTR?

Answer:

PTR-> A; // point to the operator, or you can use this operator (* PTR). A. We recommend that you use the former PTR-> B; PTR-> C;

　　

How can I access the three member variables of SS through the pointer pstr?

Answer:

* Pstr; // access a, a member of the SS. * (Pstr + 1); // access B, a member of SS. * (Pstr + 2) // access the SS member C.

Although I have adjusted the above Code on my msvc ++ 6.0, It is not formal to use pstr to access structure members. To explain why it is not formal, let's take a look at how to access each unit of the array through a pointer: (replace the struct with an array)

Example 13:

Int array [3] = {35, 56, 37}; int * pA = array; Use Pointer Pa to access the three elements of array: * pA; // access to unit 0th * (PA + 1); // access to unit 1st * (PA + 2); // access to unit 2nd

The format is the same as the format of the informal method used to access structure members through pointers.

When all C/C ++ compilers arrange the elements of an array, they always store each array unit in a continuous storage area. There is no gap between the Unit and the unit. However, when each member of the structure object is stored, in a compiling environment, it may need word alignment, double-word alignment, or other alignment, you need to add several "fill bytes" between two adjacent members, which may lead to several gaps between each member.

Therefore, in example 12, even if * pstr accesses the first member variable A of the structure object SS, * (pstr + 1) cannot be guaranteed to access structure member B. Because there may be several Padding Bytes between member A and member B, maybe * (pstr + 1) will access these padding bytes. This also proves the flexibility of pointers. If your goal is to see whether there are any bytes filled with each structure member, hey, this is a good method.

However, the correct method for accessing the structure members with pointers is to use the pointer PTR method in example 12.

7. Relationship between pointers and functions

You can declare a pointer as a pointer to a function.

Int fun1 (char *, INT); int (* pfun1) (char *, INT); pfun1 = fun1; int A = (* pfun1) ("abcdefg", 7 ); // call a function through a function pointer. You can use pointers as function parameters. In a function call statement, you can use a pointer expression as a real parameter. Example 14: int fun (char *); INTA; char STR [] = "abcdefghijklmn"; A = fun (STR); int fun (char * s) {int num = 0; For (INT I = 0;) {num + = * s; s ++;} return num ;}

In this example, function fun counts the sum of the ASCII values of each character in a string. As mentioned above, the array name is also a pointer. In function calls, after STR is passed to the form parameter S as a real parameter, the STR value is actually passed to S, s points to the same address as STR points to, But STR and S occupy their own storage space. In the function body, the auto-increment 1 operation on S does not mean that the STR auto-increment 1 operation is performed at the same time.

8. pointer type conversion

When we initialize a pointer or assign a value to a pointer, the left side of the value pair is a pointer, and the right side of the value pair is a pointer expression. In most cases, the pointer type is the same as that of the pointer expression, and the pointer type is the same as that of the pointer expression.

Example 15:

Float F = 12.3; float * fptr = & F; int * P; In the above example, what should we do if we want the pointer P to point to the real number f? Is the following statement used? P = & F;

No. Because the pointer P is of the int type *, it points to the int type. The result of Expression & F is a pointer. the pointer type is float *, and the pointer type is float. The two are inconsistent. The direct assignment method is not feasible. At least in my msvc ++ 6.0, the assignment statement for the pointer requires that the types on both sides of the assignment number be consistent, and the types to be pointed to are also consistent. I have not tried it on other compilers, you can try it. To achieve our goal, we need to perform "forced type conversion ":

P = (int *) & F;

If there is a pointer P, we need to change its type and the type

Tyep * type, the syntax format is: (type *) P;

In this way, the forced type conversion result is a new pointer, and the type of the new pointer is

Type *. It points to type, and the address it points to is the address pointed to by the original pointer. All properties of the original pointer P are not modified. (Remember)

If a function uses a pointer as a form parameter, the type must be consistent during the combination of the real parameters and form parameters in the function call statement. Otherwise, a forced conversion is required.

Example 16:

void fun(char*); int a=125,b; fun((char*)&a); void fun(char*s) { charc; c=*(s+3);*(s+3)=*(s+0);*(s+0)=c; c=*(s+2);*(s+2)=*(s+1);*(s+1)=c; } 

Note that this is a 32-bit program, so the int type occupies four bytes, And the char type occupies one byte. Function fun is used to reverse the order of four bytes of an integer. Have you noticed? In a function call statement, the result of the real parameter & A is a pointer. Its type is int *, and its type is int. The pointer type of the parameter is char *, which points to Char. In this way, in the combination of real parameters and form parameters, we must perform a conversion from the int * type to the char * type. Using this example, we can imagine the conversion process of the compiler: the compiler first constructs a temporary pointer char * temp, and then runs temp = (char *) &, finally, pass the Temp value to S. So the final result is: the type of S is char *, which points to Char, and the address it points to is the first address of.

We already know that the pointer value is the address pointed to by the pointer. In a 32-bit program, the pointer value is actually a 32-bit integer. Can an integer be directly assigned to the pointer as the pointer value? Like the following statement:

Unsigned int A; type * PTR; // type: int, Char, or structure type. A = 20345686; PTR = 20345686; // Our goal is to point the pointer PTR to the address 20345686ptr =; // Our goal is to make the pointer PTR point to address 20345686 and compile it. The result shows that the last two statements are all incorrect. So our goal cannot be achieved? No, there is another way: Unsigned int A; type * PTR; // type is int, Char, or structure type. A = N // n must represent a valid address; PTR = (type *) A; // Haha, that's all.

Strictly speaking, the type * here is different from the type * in the pointer type conversion. Here (type *) means to treat the value of the unsigned integer a as an address. It is emphasized that the value of a must represent a valid address. Otherwise, an invalid operation error occurs when you use PTR.

Think about whether or not we can take the address pointed to by the pointer as an integer. Yes. The following example shows how to take the value of a pointer as an integer and then assign this Integer as an address to a pointer:

Example 17:

Int A = 123, B; int * PTR = & A; char * STR; B = (INT) PTR; // take the value of the pointer PTR as an integer. STR = (char *) B; // assign the value of this integer to the pointer STR as an address.

Now we know that we can take the pointer value as an integer, or assign an integer as an address to a pointer.

9. pointer security issues

See the following example:

Example 18:

char s='a'; int *ptr; ptr=(int *)&s; *ptr=1298； 

Pointer PTR is an int * type pointer, which points to the int type. It points to the first address of S. In 32-bit programs, s occupies one byte, And the int type occupies four bytes. The last statement not only changes the byte occupied by S, but also changes the three bytes in the high address direction adjacent to S. What are these three bytes? Only the compiled program knows, but the program writer is unlikely to know. Maybe these three bytes store very important data, maybe these three bytes are exactly a piece of code of the program, and because of your sloppy application of pointers, the value of these three bytes has been changed! This will cause a collapse error.

Let's take another example:

Example 19:

char a; int *ptr=&a; ptr++; *ptr=115; 

This example can be fully compiled and executed. But do you see? After we perform the auto-increment 1 operation on the pointer PTR, the PTR points to a storage area adjacent to the high address of the shaping variable. What is in this bucket? We don't know. It may be a very important piece of data, or even a piece of code. The first sentence is to write a piece of data into this bucket! This is a serious error. So when using pointers, the programmer must be very clear about where my pointers actually point. When using pointers to access arrays, be sure not to go beyond the low-end and high-end limits of arrays. Otherwise, similar errors may occur.

In the forced type conversion of pointer: ptr1 = (type *) ptr2, if sizeof (ptr2 type) is greater than sizeof (ptr1 type ), therefore, it is safe to use the pointer ptr1 to access the storage zone pointed to by ptr2. If sizeof (ptr2 type) is smaller than sizeof (ptr1 type), it is insecure when ptr1 is used to access the bucket to which ptr2 points. As for why, readers should think about it in combination with example 18.

10. Conclusion

Now, do you think pointer is no longer as scared as you think? If your answer is: yes, I'm not afraid! Haha, congratulations, you have mastered the essence of C language. The only difficulty in C is Pointer. It's just about getting a pointer to other dishes. What's important is practice. Okay, let's pause the c journey and start our c ++ programming. C is a language that is very convenient for underlying operations. However, it is not convenient to develop large programs, at least it is not easy to maintain. In addition, C ++ is an object-oriented language and is now basically an object-oriented language. Therefore, we recommend that you learn C ++. C ++ is a difficult language to learn and use. It is not that easy to grasp C ++. after learning the basics, you should learn the data structure. algorithms are always the same, programming languages are endless. After learning, take the STL bone seriously. We recommend books -------- model programming and STL source code analysis. If you meet this requirement, congratulations again. You are already a programmer or even an algorithm expert, because STL has a lot of excellent and efficient algorithms. Alas, it's time to say goodbye. Let's write our life together in our language. Finally, laugh, haha, and go to bed. So tired, it's all