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#include <stdio.h>
void main() {
	char* ps = "hello bit";// In essence, the address of the first character of the string is stored in ps The pointer 
	char arr[] = "hello bit";// Essentially, put the string into an array arr in , however arr The array name represents the address of the first element 
	printf("%c\n", *ps);//h
	printf("%c\n", *arr);//h
	// The address of the starting position can be printed , Because it prints in string form , So I will look for it from the back “\0”
	printf("%s\n", arr);
	printf("%s\n", ps);
}

#include <stdio.h>
void main() {
	char str1[] = "hello bit";// Array 1
	char str2[] = "hello bit";// Array 2
	*str1 = 'p';
	printf("%s\n", str1);// You can change 
	// The following string is a constant string. Only one copy is saved in memory and cannot be changed 
	char* str3 = "hello bit";
	char* str4 = "hello bit";
	//*str3 = 'p';
	//printf("%s\n", str3);// Non modifiable 
	printf("%p\n", str1);//0019F88C
	printf("%p\n", str2);//0019F878
	// The following addresses are different from the above two addresses , Separate a space , Is the constant string address 
	printf("%p\n", str3);//00257B30
	printf("%p\n", str4);//00257B30
}

Pointer array

Definition ： An array of pointers

#include <stdio.h>
void main() {
	int a = 10, b = 20, c = 30;
	int* arr1[3] = { &a,&b,&c };// An array of plastic pointers 
	int i = 0;
	for (i = 0; i < 3; i++) {
		printf("%d\n", *(arr1[i]));//*(arr1[i]) Or you could write it as *(*(arr1+i))
	}
}

#include <stdio.h>
void main() {
	int a[] = { 1,2,3,4,5 };
	int b[] = { 2,3,4,5,6 };
	int c[] = { 3,4,5,6,7 };
	int* arr[] = {a,b,c};
	int i = 0;
	for (i = 0; i < 3; i++) {
		int j = 0;
		for (j = 0; j < 5; j++) {
			printf("%d ", *(arr[i] + j));//*(arr[i] + j) Or you could write it as arr[i][j]
		}
		printf("\n");
	}
}

arr[i] Equivalent to ：*(arr+i)

Array pointer

Shaping the pointer ： Pointer to the shape

Character pointer ： Pointer to character

Array pointer ： Pointer to array

#include <stdio.h>
void main() {
	int arr[] = { 1,2,3,4,5 };
	//arr: Address of the first element of the array ——arr[0] The address of 
	//&arr： The address of the entire array 
	int(*parr)[10] = &arr;//parr Is an array pointer , It stores the address of the array 
	double* d[5];// Pointer array 
	double* (*p)[5] = &d;// Array pointer of pointer array 
}

& Array name and array name

#include <stdio.h>
void main() {
	int arr[10] = { 0 };
	int* p1 = arr;
	int (*p2)[10] = &arr;
	printf("%p\n", arr);
	printf("%p\n", &arr);
	printf("%p\n", arr+1);// There are more addresses than above 4 byte 
	printf("%p\n", &arr+1);// There are more addresses than above 40 byte 
	// From this point of view , The step size of the array address is several times of the element size of the address step size of the first element of the array 
}

#include <stdio.h>
void main() {
	int arr[10] = {1,2,3,4,5,6,7,8,9,10};
	int(*p)[10] = &arr;
	int i = 0;
	for (i = 0; i < 10; i++) {
		printf("%d ", *((*p) + i));//p Is the address of the entire array ,*p representative arr
	}
}

Be careful ： The array name represents the address of the first element , The first element of a two-dimensional array is the first line , That is, the address of the first row of one-dimensional array, that is, the pointer of a one bit array

#include <stdio.h>
print(int(*p)[5], int r, int c) {
	int i = 0, j = 0;
	for (i = 0; i < r; i++) {
		for (j = 0; j < c; j++) {
			printf("%d ", *(*(p + i) + j));
//p+i——i+1 The address of the row array ,*(p+i)——i+1 Array of rows ,*(p+i)+j——i+1 Row array No j+1 The address of the line element ,*(*(p + i) + j)——i+1 Row array No j+1 Row element 
		}
		printf("\n");
	}
}
void main() {
	int arr[3][5] = { {1,2,3,4,5},{2,3,4,5,6},{3,4,5,6,7} };
	print(arr, 3, 5);
}

About the meaning of array code

int arr[5];// Shape array 5 Elements , Each element is int

int* p1[10];// Pointer array can store 10 individual int Pointer to type 

int(*p2)[10];//int type 10 Pointer to an array of elements 

int(*p3[10])[5];// Store array pointers ( The pointer points to int type 5 Array of elements ) Array of ( This array can store 10 Array pointers )

Array parameters and pointer parameters

One dimensional array parameters

1. There's a 10 An integer array of elements , Accept with an automatic length shaping array ——ok

2. There's a 10 An integer array of elements , use 10 An integer array of elements accepts ——ok

3. Pass an array name （int Type first element address ） use int Type pointer accepts ——ok

4. Pass an array of integer pointers that can hold 20 Elements , use 20 Element shaping pointer array accepts ——ok

5. Pass an array name （int* Type first element address ）, Accept with an integer secondary pointer ——ok

Two dimensional array parameters

About the former 3 From the following notes 1,3 all ok

4. Two dimensional array parameters , Its array name represents the address of the first element, that is, the address of the first row of one-dimensional array , From this point of view , Only 6ok

Be careful ： Only dual addresses can be accepted with secondary pointers , When accepting, you have to see whether the type is right .

First level pointer parameter transfer

#include <stdio.h>
void print(int* p,int sz) {
	int i = 0;
	for (i = 0; i < sz; i++) {
		printf("%d ", *(p + i));
	}
}
void main() {
	int arr1[10] = { 1,2,3,4,5,6,7,8,9,10 };
	int* p = arr1;
	int sz = sizeof(arr1) / sizeof(arr1[0]);
	print(p, sz);
}

The secondary pointer transmits parameters

#include <stdio.h>
void test(int** p) {
	**p = 20;
}
void main() {
	int a = 10;
	int* pa = &a;
	int** ppa = &pa;
	test(ppa);
	int b = 7;
	int* arr[10] = { &b };
	test(arr);
	printf("%d\n", a);
	printf("%d\n", *(arr[0]));
}

A function pointer

Definition ： Pointer to function （ A pointer to the address of the function ）

#include <stdio.h>
int Add(int a,int b) {
	return a + b;
}
void main() {
	//& Function name —— Get the address of the function 
	printf("%p\n", &Add);// From this point of view , You can get the address of the function 
	// Be careful ： Array name ！=& Array name , But the function name ==& Function name （ Function has no address of the first element ）
	printf("%p\n", Add);
	// Save function pointer 
	int (*p1)(int, int) = &Add;// Of course , Because the function name ==& Function name , So the & Omission 
	printf("%p\n", p1);
	int (*p2)(int, int) = Add;
	printf("%p\n", p2);
	// Function call 
	int ret=(*p1)(3, 5);
	int res=p2(3, 5);// Call directly with the function name ——p2==&Add and &Add==Add therefore p2==Add
	printf("%d\n",ret);
	printf("%d\n",res);
}

Function pointer type ： return type (*)( Parameter type list )

Code reading

Declaration that the function pointer type is the return value

void (* signal(int,void(*)(int)))(int);
replace ——
void(*)(int) signal(int,void(*)(int));

// Easy to write 
typedef void(*pfun_t)(int);// Yes void(*)(int) Rename the function pointer type to pfun_t;
pfun_t signal(int, pfun_t);

Function pointer array

Definition ： An array of function pointers

#include <stdio.h>
int Add(int a,int b) {
	return a + b;
}
int Sub(int a, int b) {
	return a - b;
}
void main() {
	int (*p1)(int, int) = Add;
	int (*p2)(int, int) = Sub;
	int (*parr[2])(int, int) = {p1,p2};// Function pointer array can store 2 The return values are int parameter list 2 individual int Function pointer of 
	int ret = (*parr[1])(8, 5);// Function call through function pointer array 1
	int res = parr[0](8, 5);// Function call through function pointer array 2
	printf("%d\n", ret);
	printf("%d\n", res);
}

A pointer to an array of function pointers

Definition ： Get the address of the function pointer array

    int arr1[5];// Shape array 
	int (*p1)[5]=&arr1;// A pointer to an integer array 
	int* arr2[5];// An array of integer pointers 
	int* (*p2)[5] = &arr2;// Integer pointer array pointer 
	int (*p3)(int,int);// A function pointer 
	int (*p4[5])(int, int);// An array of function pointers 
	int (*(*p4)[5])(int, int);// Function pointer array pointer

Callback function

Definition ：

give an example ：b The function has a Function pointer as parameter , When passed a The pointer of the function calls a function , This mechanism is called callback function mechanism , The function called by the pointer in turn is called a callback function .

#include <stdio.h>
int Add(int a,int b) {
	return a + b;
}
int AP(int(*p)(int,int)) {
	return p(2, 3);// Here I have directly omitted *
}
void main() {
	int(*ap)(int(*)(int, int)) = &AP;
	int ret=ap(&Add);
	printf("%d\n", ret);
	return;
}

#include <stdio.h>
int Add(int a,int b) {
	return a + b;
}
int AP(int(*p)(int,int)) {
	return p(2, 3);// Here I have directly omitted *
}
void main() {
	// How to write it 2—— Anyway &AP==ap==AP==*ap So will *ap use AP replace 
	int ret=AP(&Add);
	printf("%d\n", ret);
	return;
}

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