Pointer for in-depth analysis (problem solution)

  Preface

do person ： Little snail rushes forward

quotes ： I can accept failure , But I can't accept giving up

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Catalog

One dimensional array  

  A character array

  Constant string

  Two dimensional array

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about C Language pointers are very important , In order to understand pointer more thoroughly , Now I'll share some pointers , Play with the pointer in the analysis of the topic .

Before doing the problem , We need to understand what array names are .

Array name

Usually refers to the address of the first element

  In two special cases , Refers to the address of the entire array

1 & Address array name

2 sizeof( Array name )（ alone ）

One dimensional array  

// One dimensional array 
	int a[] = { 1,2,3,4 };
	printf("%d\n", sizeof(a));//a Refers to the entire array name , So the byte size is 4*4=16
	printf("%d\n", sizeof(a + 0));//a Refers to the address of the first element ,a+0 Indicates the number of bytes skipped 4*0=0, Point to 1, So the byte size is 4
	printf("%d\n", sizeof(*a));//a Refers to the address of the first element , Yes *a find 1,1 The byte size of is 4
	printf("%d\n", sizeof(a + 1));//a Refers to the address of the first element ,a+1 Indicates the number of bytes skipped 4*1=4, Point to 2, So the number of bytes is 4
	printf("%d\n", sizeof(a[1]));//a[1]-->*(a+1), find 2, So the number of bytes is 4
	printf("%d\n", sizeof(&a));//&a Represents the address of the whole array , The address in 32 The size of each platform is 4 Bytes 
	printf("%d\n", sizeof(*&a));//&a Represents the address of the whole array ,*&a Find each element of the array , Byte size 4*4=16
	printf("%d\n", sizeof(&a + 1));//&a+1, It still indicates that the size of the address is 4 Bytes 
	printf("%d\n", sizeof(&a[0]));//&a[0]-->&*(a+0), Express &1, Take out 1 The address of , The size of the address is still 4 Bytes 
	printf("%d\n", sizeof(&a[0] + 1));//&a[0]+1 Express &2, The byte size is 4

For one-dimensional arrays , We must always remember , What is the meaning of array .

stay 32 Bit platform Next The size of the address is always 4 Bytes size

  A character array

There are two initial situations for character arrays , Now we will experience the topic in detail .

situation 1

sizeof Question type

	char arr[] = { 'a','b','c','d','e','f' };
	printf("%d\n", sizeof(arr));// The size of the entire array , Each element of the array is char type , The size is 1 Bytes , therefore 1*6=6
	printf("%d\n", sizeof(arr + 0));//arr+0 Refers to the address of the first element , therefore 4 Bytes 
	printf("%d\n", sizeof(*arr));//*arr find 'a', therefore 1 Bytes 
	printf("%d\n", sizeof(arr[1]));//arr[1]-->*(arr+1), find 'b', therefore 1 Bytes 
	printf("%d\n", sizeof(&arr));//&arr Represents the address of the whole array , So the byte size is 4 Bytes 
	printf("%d\n", sizeof(&arr + 1));//&arr+1 It means to get the address after the array , therefore 4 Bytes 
	printf("%d\n", sizeof(&arr[0] + 1));// Empathy , therefore 4 Bytes 

strlen Question type

	char arr[] = { 'a','b','c','d','e','f' };
	printf("%d\n", strlen(arr));//'\0' The location is undetermined , Random value 
	printf("%d\n", strlen(arr + 0));// Random value 
	printf("%d\n", strlen(*arr));// There is a problem with the transmission , Unable to evaluate -->strlen(a)-->strlen(97)( Wild pointer )
	printf("%d\n", strlen(arr[1]));//arr[1]-->*(arr+1), There is a problem with the transmission , Unable to evaluate -->strlen(b)-->strlen(98)
	printf("%d\n", strlen(&arr));//&arr The address of the entire array , Random value 
	printf("%d\n", strlen(&arr + 1));// Random value -1*6
	printf("%d\n", strlen(&arr[0] + 1));//&*arr[0]+1--&*(arr+0)+1, Random value -1

inductive

sizeof It's an operator, which asks for the size of bytes ,strlen Is the function of the size of the string .

char arr[] = { 'a','b','c','d','e','f' };// This initialization method does not add '\0'.

arr[0] Can be equivalent to *(arr+0)

  situation 2

sizeof Question type

char arr[] = "abcdef";
	printf("%d\n", sizeof(arr));// Find the size of the entire array , So bytes 7
	printf("%d\n", sizeof(arr + 0));// Find the size of the address of the first element , So bytes 4
	printf("%d\n", sizeof(*arr));// Find the size of the first element , So bytes 1
	printf("%d\n", sizeof(arr[1]));//arr[1]-->*(arr+1), seek b Size , So bytes 1
	printf("%d\n", sizeof(&arr));// Find the size of the entire array address , So bytes 4
	printf("%d\n", sizeof(&arr + 1));// Skip the size of the entire array address , So bytes 4
	printf("%d\n", sizeof(&arr[0] + 1));//&arr[0]+1-->&*(arr+0)+1, So bytes 4

strlen Question type

    char arr[] = "abcdef";  
    printf("%d\n", strlen(arr));// Pass the address of the first element of the array , So there is 6 Characters 
	printf("%d\n", strlen(arr + 0));// ditto ,6 Characters 
	printf("%d\n", strlen(*arr));// There is a problem with the transmission , No calculation 
	printf("%d\n", strlen(arr[1]));// There is a problem with the transmission , No calculation 
	printf("%d\n", strlen(&arr));// Pass the address of the whole array （ The same value as the address of the first element ）, therefore 6 Characters 
	printf("%d\n", strlen(&arr + 1));// Skip array , After '\0' The location of is uncertain , So it's a random value 
	printf("%d\n", strlen(&arr[0] + 1));// Pass on b The address of , therefore 5 Characters 

inductive

 char arr[] = "abcdef";// There is... At the end of the character '\0'

  Constant string

sizeof Question type

	char* p = "abcdef";
	//printf("%d\n", sizeof(p));//a The address of , So bytes 4
	//printf("%d\n", sizeof(p + 1));//b The address of , So bytes 4
	//printf("%d\n", sizeof(*p));// character a, So bytes 1
	//printf("%d\n", sizeof(p[0]));//p[0]-->*(p+0), character a, So bytes 1
	//printf("%d\n", sizeof(&p));// Get the address of the entire constant string , So bytes 4
	//printf("%d\n", sizeof(&p + 1));// Point to the address after the constant character , So bytes 4
	//printf("%d\n", sizeof(&p[0] + 1));//&*(p+0)+1-->&b, So bytes 4

strlen Question type  

    char* p = "abcdef";
	printf("%d\n", strlen(p));//6 Give character 
	printf("%d\n", strlen(p + 1));// Pass on b Your address used to , character 5
	printf("%d\n", strlen(*p));// There is a problem with the transmission , No calculation 
	printf("%d\n", strlen(p[0]));// There is a problem with the transmission , No calculation 
	printf("%d\n", strlen(&p));// hold a Send the address of the address , meet '\0' The location of is uncertain , Random value 
	printf("%d\n", strlen(&p + 1));// hold b Send the address of the address , meet '\0' The location of is uncertain , Random value 
	printf("%d\n", strlen(&p[0] + 1));// Pass on b The address of , therefore 5 Characters 

inductive

Variable name of constant string Is the first address of the constant string .

&p about strlen In terms of functions , What passed is a secondary pointer , First of all, the address of the first address is stored in memory. I don't know if it is Big end storage still Small end storage , This makes you encounter '\0' The position of cannot be judged , So use strlen The result is a random value .

  Two dimensional array

	int a[3][4] = { 0 };
	printf("%d\n", sizeof(a));//a Indicates that the size of the entire array is calculated ,48
	printf("%d\n", sizeof(a[0][0]));// Calculate the size of the first element ,4
	printf("%d\n", sizeof(a[0]));//a[0]-->*(a+0), Indicates the size of the first row of the count 16
	printf("%d\n", sizeof(a[0] + 1));// Calculate the second line 1 Size of element addresses 4
	printf("%d\n", sizeof(*(a[0] + 1)));// Count the second line 1 The size of an element 4
	printf("%d\n", sizeof(a + 1));// Calculate the second line 1 Size of element addresses 4
	printf("%d\n", sizeof(*(a + 1)));// Calculate the size of the second line 16
	printf("%d\n", sizeof(&a[0] + 1));// Calculate the size of the address in the second line 4
	printf("%d\n", sizeof(*(&a[0] + 1)));// Calculate the size of the second line 16
	printf("%d\n", sizeof(*a));// Calculate the size of the first row 16
	printf("%d\n", sizeof(a[3]));// Although the array is out of bounds , But what we ask is that the size of bytes is ok , It is equivalent to finding the number of arrays N The address of the line ,16

inductive

The array name of the two-dimensional array points to the... Of the array 1 That's ok

summary

The meaning of array names ：

1. sizeof( Array name ), The array name here represents the entire array , It calculates the size of the entire array .

2. & Array name , The array name here represents the entire array , It takes out the address of the entire array .

3. In addition, all array names represent the address of the first element .

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