C language -- 22 one dimensional array

1、 One dimensional array creation and initialization

1.1 Array creation
An array is a collection of elements of the same type . How to create an array ：

type_t arr_name    [const_n]
//type_t   Is the element type of the exponential group 
//condt_n   Is a constant expression , Used to specify the size of an array 

Example of array creation ：

     // The basic creation form of array 
	// Element type   Array name   Element number 
	int arr[8];// Array name is smaller arr,[8] It means you want to put 8 Elements , this 8 Elements. What type is each element ？ yes int
	char ch[5];// Array name ch,[5] It means to put 5 Elements , this 5 The types of the elements are all char type 

Error creating ： Variables are used in the creation process
c99 Syntax supports variable length arrays （ The size of the array is a variable ）, however vs2022 Variable length arrays are not supported, so an error will be reported .

Array creation ,[] You have to give a constant in the , You can't use variables
1.2 Initialization of an array
The initialization of array refers to giving some reasonable initial values to the contents of the array while creating the array （ initialization ）. Such as ：int a = 10; Is to initialize the array . For initialization of arrays , It is divided into complete initialization and incomplete initialization

• Fully initialized

int main()
{
    
	int a = 10;// initialization 
	int arr[10] = {
    1,2,3,4,5,6,7,8,9,10};// Fully initialized 
	return 0;
}

• Incomplete initialization

int arr[10] = {
     1,2,3,4,5 };// Incomplete initialization 

There is one caveat , If in the definition of array , Do not specify the number of elements in the array , Then the array will determine the number of elements according to the number in the subsequent element set . Such as int arr2[] = {1,2,3,4,5};, The number of element sets is 5, So the array size defaults to 5.
Take a look at the following array definitions ：
If you want to create an array without specifying the size of the array, you have to initialize it . The number of elements in the array is determined according to the contents of initialization . But for the following code, we should distinguish how to allocate memory

char ch1[5] = {
    'h','i'};
char ch2[] = {
     'h','i' };
// Use string to initialize 
char ch3[5] = "hi";// The first three elements of this array will be placed "h"、"i"、"\0", What remains is 0（ It can also be understood as \0,, because \0 Of ASCII The code value is also 0）
char ch4[] = "hi";// But this is the space created according to the string content , So there will be no extra , It will only be "h"、"i"、"\0"

About the difference between the two ：
a、 The difference between content printing ：

char ch2[] = {
     'h','i' };// This is put in "h"、"i" Two elements 
char ch4[] = "hi";// This will be put in "h"、"i"、"\0" Three elements 

char ch4[] = "hi"; It means to give ch4 Opened up a space , Deposit "h"、“i”、“\0" Three elements , But what is in front of these three elements , What is behind these three elements , These we don't know .
char ch2[] = { 'h','i' }; It refers to placing in the open space "h”、“i”, At this time, I still don't know what is placed in front of these two elements , I don't know what is placed behind , So when printing the contents of the array, you will find the terminator "\0", End printing whenever you find it .
b、 The difference in length
Add these two sentences of code to verify ：

char ch2[] = {
     'h','i' };// This is put in "h"、"i" Two elements 
char ch4[] = "hi";// This will be put in "h"、"i"、"\0" Three elements 
printf("%d\n", strlen(ch2));// Random value 
printf("%d\n", strlen(ch4));// The length is 2

2、 The use of one-dimensional arrays

The use of arrays involves an operator [ ]---- Subscript reference operator , It's actually an array access operator .

int main()
{
    
	int arr[10] = {
     0 };// Incomplete initialization , The first is 0, The remaining elements are all 0; This initializes all ten elements to 0 It's the same 
	arr[4] = 5;// Mark the subscript as 4 The element value of is changed to 5
	int i = 0;
	int sz = sizeof(arr) / sizeof(arr[0]);// Find the length of the array 
	for (i = 0; i < sz; i++)
	{
    
		printf("%d ",arr[i]);
	}
	return 0;
}

The code runs as follows ：

0 0 0 0 5 0 0 0 0 0

summary ：

• Arrays are accessed through subscripts , The subscript is from 0 Start
• The size of the array can be calculated

3、 One dimensional array storage in memory

int main()
{
    
	int arr[10] = {
     0 };
	arr[4] = 5;
	int i = 0;
	int sz = sizeof(arr) / sizeof(arr[0]);// Find the length of the array 
	for (i = 0; i < sz; i++)
	{
    
		printf("&arr[%d]=%p\n",i, &arr[i]);
		//%p It is printed in the format of address --- Hexadecimal printing （ Will make up the number ）
	}
	return 0;
}

&arr[0]=000000CA3CB3F8C8
&arr[1]=000000CA3CB3F8CC
&arr[2]=000000CA3CB3F8D0
&arr[3]=000000CA3CB3F8D4
&arr[4]=000000CA3CB3F8D8
&arr[5]=000000CA3CB3F8DC
&arr[6]=000000CA3CB3F8E0
&arr[7]=000000CA3CB3F8E4
&arr[8]=000000CA3CB3F8E8
&arr[9]=000000CA3CB3F8EC

The difference between each address and each address 4 Bytes , Because every integer element is 4 Bytes , So the difference between two integer elements must be four bytes . From this code and its running results, we can see ： One dimensional arrays are stored continuously in memory ; As the array subscript grows , The address changes from low to high ; The array name is the address of the first element of the array
The difference between hexadecimal printing and address printing ：

printf("%x\n",0x12);
printf("%p\n",0x12);

12
0000000000000012

Because arrays are stored continuously in memory , Therefore, you can access the address of the first element later, and then get each of the following elements one by one . The code is as follows ：

int main()
{
    
	int arr[10] = {
     1,2,3,4,5,6,7,8,9,10};
	int* p = arr;// The array name is the address of the first element of the array , It's the address , So you can put it in the pointer ;
	int i = 0;
	for (i = 0; i < 10; i++)
	{
    
		printf("%d ",*p);
		p++;
	}
	return 0;
}

The code runs as follows ：

1 2 3 4 5 6 7 8 9 10