C primer plus learning notes - 5. Pointer

What is a pointer

A pointer is a variable whose value is a memory address .
Analogy ：int The value of a type variable is an integer . The value of a pointer type variable is an address .

int a = 4;//  Variable a It's an integer 
int * p = 0x3434523; // Variable p It's an address value 

But the value of this address is generally unknown to us , Generally, it is handed over to the system after allocation , And then we use

int a = 4;
int *p = &a; // Because the system has given variables a Assigned an address , So we can get a The address of is then assigned to p

Here the pointer is p, We can also say pointer p Point to variable a.

* Operator

Operator * We also call it dereference operator .

Declare a pointer ：

int *p; //p It's pointing int Pointer to type variable .
char *pc; //pc It's a point char Pointer to type 

here , asterisk * Express Statement It means , The declared variable is a pointer . It's a statement p It's a pointer , This pointer points to int Type variable . In other words, stored in p The data type on the address is int.
pc The value pointed to is (*pc) yes char type .

int a = 4;
int *p = &a; 
printf(%d, *p); //* Will find the address p Values that exist on 

Pointer types

Pointer is actually a new type , He is not an integer type , It is the pointer type . such as p, Its type is pointer .
int*p and char*pc in p and pc All are Pointer types , But the types of values stored on their addresses are different , One stored is int type , One stored is char type .
Not that the type of this pointer is char, Another pointer type is int. It should be said that they are all pointer types , The types of values pointed to are different .

int main(void)
{
    
    int a = 1000;
    int *p = &a; 
    printf("%p \n", p); // Print address  0x7ffeba874aac
    printf("%d \n", *p);  // 1000
    char *pc = &a;
    printf("%p \n", pc); // Print address  0x7ffeba874aac
    printf("%d \n", *pc); // -24

    printf("%p \n", p+1); //0x7ffeba874ab0
    printf("%p \n", pc+1); //0x7ffeba874aad
    
    return 0;
}

Embodiment is based on what ： For example, in the above example, if int In units of , We print p The address of , And get p Place the value of the , With int Type fetch , Namely 4 Bytes to get 1000.
When we pc Pointer get a When the address of , Or take the value of this position , But at this time, we are char The type is taken in units , Then the value obtained is the value of one byte -24.

Let's look at the address +1, such as p+1 In the memory address, you will use int To move the smallest unit , Move 4 byte , That is, the address value will increase 4, and pc+1 Then only move 1 The byte address value only increases 1.

Personal understanding

The book does not say that pointer types are different , But I think it can be understood as different types of pointers , such as int Pointer to type , and char Pointer to type , These two pointers belong to different types , They move in their own units , And the values pointed to are of different types . There is nothing wrong .

The value of the address is forced

The address is represented by an unsigned integer . We can use uint_64t
I think an example can better understand pointer , Is to take a uint_64 The number of type is strongly converted to pointer type .

int main(void)
{
    
    int a = 1000;
    int *p = &a; 
    printf("%p \n", p); // Print address  0x7ffeba874aac

    uint64_t val_p = (uint64_t)p; //p It's a pointer type , Let's give him a strong turn 
    printf("%lx \n", val_p); // 7ffeba874aac
   
    // &a  Is the value of an address , We can force it into a pointer type 
    
    printf("%lx \n", &a); // &a  The value of is 7ffeba874aac
    int *ip;
    ip = (int *) &a; // such ip This is the address 
    printf("%p \n",  ip); //0x7ffeba874aac
    printf("%d \n", *ip); // 1000  We can get the address 

    int *ip2;
    ip2 = (void *) &a; //z This address can be strongly converted to any pointer type 
    printf("%p \n",  ip2); //0x7ffeba874aac
    printf("%d \n", *ip2); // 1000  But when printing, we need to specify the type of the printed variable 

	printf("%zd \n", sizeof(int*)); //8
    printf("%zd \n", sizeof(ip)); //8
    printf("%zd \n", sizeof(void *));//8
    return 0;
}

The size of pointer types is 8 byte .