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When defining character pointers , There are two ways of writing ：
	// The first one is 
    char ch[] = "hello";// Define a constant string 
	char* ptr = &ch; // Take out ch The address of , Save to character pointer ptr in 
    // The second kind 
	char* str = "hello";// Direct the character pointer to the constant string 
Maybe when touching the character pointer , Have had such an idea —— Since the string is stored in a pointer variable , Then the dereference operation should be output as is , But that's not the case , Its essence is to store the address of the first character of the string in the pointer variable , The pointer can find this contiguous space and access it through this address ;

Answer the question with the following code ：

#include <stdio.h>
int main() 
{
	char ch[] = "hello";
	char* ptr = &ch; 
	char* str = "hello";
	printf("%c\n", *ptr);
	printf("%c\n", *(ptr+1));
	printf("%c\n", *str);
	printf("%c\n", *(str + 1));
	return 0;
}

From the results of operation , The result of dereference is h, explain ptr Pointers and str The pointer stores h The address of ,*（ptr+1） The printed result is e, The description string is also accessed by subscript （ Continuous space ）;

Two 、 Pointer array

Concept ： I have an array , The elements stored in the array are pointers （ Address ） ;

#include <stdio.h>
int main()
{
	int a = 10;
	int b = 20;
	int c = 30;
	int* arr[3] = { &a,&b,&c };// This is how the pointer array is defined 
	printf("%d\n", *arr[0]);// or **arr
    //**arr---arr Is the array name, that is, the address of the first element , After dereferencing, we get  &a , And then you can get 10
	return 0;
}

3、 ... and 、 Array pointer

3.1 Definition of array pointer

Array pointer is a pointer or an array ？（ The answer is ： The pointer ）
Compare the following two codes ：
int* p1[5];// Pointer array 
int (*p2)[5];// Array pointer （ Row pointer ）
( ) The priority of [ ] higher ; [ ] The priority of * higher ;
1. p1 With the first [ ] Combined table name p It's an array , This array has 5 Elements , The type of each element is int*;
2. p2 With the first * combination , indicate p2 Is a pointer , Points to an array , Each element group has 5 Elements , The type of each element is int;

3.2& Array name and array name

The array name indicates the address of the first element of the array , that & What does the array name mean ？

#include <stdio.h>
int main()
{
	int arr[10] = { 0 };
    printf("arr = %p\n", arr);
    printf("&arr= %p\n", &arr);
    printf("arr+1 = %p\n", arr+1);
    printf("&arr+1= %p\n", &arr+1);
	return 0;
}

3.3 The use of array pointers

Often used for two-dimensional arrays

#include <stdio.h>
int main()
{
	int arr[2][5] = { 0,1,2,3,4,5,6,7,8,9 };
	int(*parr)[5] = &arr;
	for (int i = 0; i < 2; i++)
	{
		for (int j = 0; j < 5; j++)
		{
			printf("%d ", parr[i][j]);
			//printf("%d ", (*(parr+i))[j]);
			//printf("%d ", *(parr[i] + j));
			//printf("%d ", *(*(parr + i)+j));
		}
		printf("\n");
	}
	return 0;
}

parr[ i ] [ j ] = ( * ( parr + i ) ) [ j ] ) = * ( parr [ i ] + j ) = * ( * ( parr + i ) + j );

Four 、 Parameter passing of pointer

The ginseng ： The essential meaning is that the transmitter and the receiver are of the same type or the receiver belongs to any type of acceptance （ Such as ：void）

#include <stdio.h>
void print_arr1(int arr[3][5], int row, int col)
{// This is a common two-dimensional array parameter transfer , When transferring parameters, note that rows can be omitted, but columns cannot 
	int i = 0;
	for (i = 0; i < row; i++)
	{
		for (int j = 0; j < col; j++)
		{
			printf("%d ", arr[i][j]);
		}
		printf("\n");
	}
}
void print_arr2(int(*arr)[5], int row, int col)
{// Here, the address of the first row of the two-dimensional array is passed to an array pointer 
	int i = 0;
	for (i = 0; i < row; i++)
	{
		for (int j = 0; j < col; j++)
		{
			printf("%d ", arr[i][j]);
		}
		printf("\n");
	}
}
int main()
{
	int arr[3][5] = { 1,2,3,4,5,6,7,8,9,10 };
	print_arr1(arr, 3, 5);
	// Array name arr, Represents the address of the first element 
	// But the first element of a two-dimensional array is the first row of a two-dimensional array 
	// So the message here is arr, It's actually equivalent to the address on the first line , Is the address of a one-dimensional array 
	// You can use an array pointer to receive 
	print_arr2(arr, 3, 5);
	return 0;
}}

5、 ... and 、 A function pointer

Essential meaning ： Is to store the address of the function in a pointer variable ;

5.1 Code Introduction

#include <stdio.h>
int Add(int a, int b)
{
	return a + b;

}
int main()
{
	int a = 10;
	int b = 20;
	int (*pa)(int, int) = Add;
	int ret = pa(a, b);
	printf("%d\n", ret);
	return 0;
}

int (*pa)(int, int) = Add;
Add Is the function name , It's the address of the function ,Add The type is int (int a, int b); Save it in a pointer variable , The type of this pointer variable should also be int ( * ) (int a, int b), Or you could write it as int( * ) (int , int )
It must not be written as int *pa(int, int) = Add; In this code pa It is expressed as a function , The type is int*, So the parentheses are Very important Of
int ret = pa(a, b);
pa Deposit is Add Address of function , Here we use pa To call Add function , And give it a reference ;

5.2 Feel the happiness of function pointer

5.2.1 analysis ( * ( void ( * ) ( ) ) 0 ) ( ); Code

5.2.2 analysis void ( * signal ( int , void ( * ) ( int ) ) ) ( int ); Code

6、 ... and 、 Function pointer array

meaning ： It's essentially an array , The elements in the array are function pointers ; To put it simply , There is an array containing the addresses of many functions
int (*p[5])(int,int);
p[ 5 ] Indicates an array , Its type is int ( * )( int , int );

Without touching the function pointer array , If you want to write a code to add, subtract, multiply and divide Calculator , We can take advantage of switch Statement to select function functions , The code is as follows ;

#include <stdio.h>
int add(int a, int b)
{
	return a + b;
}
int sub(int a, int b)
{
	return a - b;
}
int mul(int a, int b)
{
	return a * b;
}
int div(int a, int b)
{
	return a / b;
}
int main()
{
	int x, y;
	int input = 1;
	int ret = 0;
	do
	{
		printf("*************************\n");
		printf("****1:add       2:sub****\n");
		printf("****3:mul       4:div****\n");
		printf("*************************\n");
		printf(" Please select ：");
		scanf("%d", &input);
		switch (input)
		{
		case 1:
			printf(" Enter the operands ：");
			scanf("%d %d", &x, &y);
			ret = add(x, y);
			printf("ret = %d\n", ret);
			break;
		case 2:
			printf(" Enter the operands ：");
			scanf("%d %d", &x, &y);
			ret = sub(x, y);
			printf("ret = %d\n", ret);
			break;
		case 3:
			printf(" Enter the operands ：");
			scanf("%d %d", &x, &y);
			ret = mul(x, y);
			printf("ret = %d\n", ret);
			break;
		case 4:
			printf(" Enter the operands ：");
			scanf("%d %d", &x, &y);
			ret = div(x, y);
			printf("ret = %d\n", ret);
			break;
		case 0:
			printf(" Exit procedure \n");
			break;
		default:
			printf(" Wrong choice \n");
			break;
		}
	} while (input);
	return 0;
}

Implementation using function pointer array ：

#include <stdio.h>
int add(int a, int b)
{
	return a + b;
}
int sub(int a, int b)
{
	return a - b;
}
int mul(int a, int b)
{
	return a * b;
}
int div(int a, int b)
{
	return a / b;
}
int main()
{
	int x, y;
	int input = 1;
	int ret = 0;
	int(*p[5])(int x, int y) = { 0, add, sub, mul, div }; 
	while (input)
	{
		printf("*************************\n");
		printf("****1.add       2.sub****\n");
		printf("****3.mul       4.div****\n");
		printf("*************************\n");
		printf(" Please select ：");
		scanf("%d", &input);
		if ((input <= 4 && input >= 1))
		{
			printf(" Enter the operands ：");
			scanf("%d %d", &x, &y);
			ret = (*p[input])(x, y);// Input input Respectively corresponding to the subscript of the corresponding function in the function pointer 
		}
		else
			printf(" Incorrect input \n");
			printf("ret = %d\n", ret);
	}
return 0;
}

7、 ... and 、 A pointer to an array of function pointers

How to define ：


    // A function pointer p
    int (*p)(int x,int y);
    // An array of function pointers p1
    int (*p1[5])(int x,int y)=&p;
    // Pointer to function array p1 The pointer to p2
    int (*(*p2)[5])(int x,int y) = &p1;

#include <stdio.h>
int add(int a, int b)
{
	return a + b;
}
int main()
{

    // A function pointer p
    int (*p)(int x, int y) = add;

    // An array of function pointers p1
    int (*p1[4])(int x, int y) = { p };// Here you can also subtract 、 Multiplication 、 The address of the division function is saved in 

    // Pointer to function array p1 The pointer to p2
    int (*(*p2)[4])(int x, int y) = &p1;

    // Use the pointer to the function pointer array to add two numbers 
    int ret = (*(*p2[0]))(10, 5);

    printf("ret=%d\n", ret);
	return 0;
}

8、 ... and 、 summary

All contents of the above pointer advanced , The understanding of the pointer in this paper is mainly based on the author's own statement , There is no professional explanation , I hope that reading these two pointers can help you .

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