Advanced pointer practice

Pointer practice

One dimensional array

// One dimensional array 
int a[] = {
    1,2,3,4};
printf("%d\n",sizeof(a));【16】---- special case ： The array list is unique sizeof Internal time , It calculates the size of the entire array 
    
printf("%d\n",sizeof(a+0));【4/8】---- The array name is the address of the first element , stay a+0 It's in the array 1 The address of , because a+0 It's the address , therefore sizeof（ Address ） The size is 4/8 Bytes 
    
printf("%d\n",sizeof(*a));【4】---- The array name is the address of the first element ,* It's the first element , So we calculate the size of the first element 
    
printf("%d\n",sizeof(a+1));【4/8】---- The array name is the address of the first element ,a+1 It's in the array 2 The address of , because a+1 It's the address , therefore sizeof（ Address ） The size is 4/8 Bytes 
    
printf("%d\n",sizeof(a[1]));【4】----a[1] Is the second element in the array , So we calculate the size of the second element 

printf("%d\n",sizeof(&a));【4/8】----&a What you get is the address of the whole array , but &a It's the address , therefore sizeof（ Address ） The size is 4/8 Bytes 
    
printf("%d\n",sizeof(*&a));【16】---- Take out the address of the whole array , Dereference the address of the entire array 
     because 8a Medium a Is the address of the first element of the array ,a<==>&a[0],*&a[0] amount to a[0], This is equivalent to * and & Offset each other 
    
printf("%d\n",sizeof(&a+1));【4/8】----&a+1 What you get is an array a The address of the back position , Because it's the address , therefore sizeof（ Address ） The size is 4/8 Bytes 
    
printf("%d\n",sizeof(&a[0]));【4/8】----&a[0] What you get is the address of the first element , Because it's the address , therefore sizeof（ Address ） The size is 4/8 Bytes 
    
printf("%d\n",sizeof(&a[0]+1));【4/8】----&a[0]+1 What you get is the address of the second element , Because it's the address , therefore sizeof（ Address ） The size is 4/8 Bytes 

A character array

##  A character array 

char arr[] = {
    'a','b','c','d','e','f'};
printf("%d\n", sizeof(arr));【6】---- special case ：arr The array list is placed alone in sizeof in , What you get is the size of the entire array 

printf("%d\n", sizeof(arr+0));【4/8】---- The array name is the first element address ,arr+0 What you get is the address of the first element , Because it's the address , therefore sizeof（ Address ） The size is 4/8 Bytes 
    
printf("%d\n", sizeof(*arr));【1】---- The array name is the address of the first element ,*arr What you get is the first element ,sizeof(char) yes 1 Bytes 

printf("%d\n", sizeof(arr[1]));【1】----arr[1] Is the size of the second element 

printf("%d\n", sizeof(&arr));【4/8】---- It takes out the address of the entire array , Because it's the address , therefore sizeof（ Address ） The size is 4/8 Bytes 
    
printf("%d\n", sizeof(&arr+1));【4/8】----&arr What you get is the address of the whole array ,&arr+1 It refers to the address of the space behind the array , Because it's the address , therefore sizeof（ Address ） Get is 4/8 Bytes 
    
printf("%d\n", sizeof(&arr[0]+1));【4/8】----&arr[0] What you get is the address of the first element of the array ,&arr[0]+1 Points to the address of the second element of the array , Because it's the address , therefore sizeof（ Address ） Get is 4/8 Bytes 

    
    
     About strlen
char arr[] = {
    'a','b','c','d','e','f'};
printf("%d\n", strlen(arr));【 Random value 】 Because there is no '\0' The existence of , So when calculating the length , What you get is a random value 
       
printf("%d\n", strlen(arr+0));【 Random value 】arr Is the address of the first element of the array ,arr+0 What you get is also the address of the first element of the array , Because there is no '\0' The existence of , So when calculating the length , What you get is a random value 
       
printf("%d\n", strlen(*arr));【 Compile error 】----*arr Get the first character 'a',strlen（97）, Pass to strlen Should be the address , Dereference to get a in the future ,a The address of is a wild pointer 
    
printf("%d\n", strlen(arr[1]));【 Compile error 】----arr[1] What you get is the second character 'b'  Pass to strlen Should be the address , Dereference to get b in the future ,b The address of is a wild pointer 
    
printf("%d\n", strlen(&arr));【 Random value 】----&arr What you get is the address of the entire array , Because there is no '\0' The existence of , So when calculating the length , What you get is a random value 
    
printf("%d\n", strlen(&arr+1));【 Random value -6】----&arr What you get is the address of the entire array ,&arr+1 It refers to the address of the space behind the character array , So when calculating the length , What you get is a size after subtracting the length of the character array , That is, random value -6
    
printf("%d\n", strlen(&arr[0]+1));【 Random value -1】----&arr[0] What you get is the address of the first element of the array ,&arr[0]+1 What you get is the address of the second element , So when calculating the length , What you get is the size of the length after subtracting the first character , That is, random value -1

char arr[] = "abcdef";
printf("%d\n", sizeof(arr));【7】----arr Put it alone sizeof On the inside , It calculates the size of the entire array , because sizeof The calculation is the size of the occupied space , So it also includes '\0'  So what we get is 7
    
printf("%d\n", sizeof(arr+0));【4/8】----arr Is the address of the first element of the array ,arr+0 What you get is the address of the first element of the array , Because it's the address , therefore sizeof（ Address ） The resulting size is 4/8
    
printf("%d\n", sizeof(*arr));【1】----*arr It's the first element , the reason being that char type , So it is 1
    
printf("%d\n", sizeof(arr[1]));【1】----arr[1] Is the second element in the array , the reason being that char type , So it is 1

printf("%d\n", sizeof(&arr));【4/8】----&arr It's the size of the whole array , Because it's the address , therefore sizeof（ Address ） The resulting size is 4/8 Bytes 
    
printf("%d\n", sizeof(&arr+1));【4/8】----&arr+1 It points to the address of the space behind the array , Because it's the address , therefore sizeof（ Address ） The size obtained is 4/8 Bytes 
    
printf("%d\n", sizeof(&arr[0]+1));【4/8】----&arr[0]+1 What you get is the address of the second element , Because it's the address , therefore sizeof（ Address ） The size obtained is 4/8 Bytes 
    
printf("%d\n", strlen(arr));【6】----strlen The calculation is '\0' The number of previous elements , Yes 6 individual 
    
printf("%d\n", strlen(arr+0));【6】----strlen The calculation is '\0' The number of previous elements , Yes 6 individual 

printf("%d\n", strlen(*arr));【 Compile error 】*arr Get the first character 'a',strlen（97）, Pass to strlen Should be the address , Dereference to get a in the future ,a The address of is a wild pointer 
    
printf("%d\n", strlen(arr[1]));【 Compile error 】----arr[1] What you get is the second character 'b'  Pass to strlen Should be the address , Dereference to get b in the future ,b The address of is a wild pointer 
    
printf("%d\n", strlen(&arr));【6】----strlen The calculation is '\0' The number of previous elements , Yes 6 individual 
    
printf("%d\n", strlen(&arr+1));【 Random value 】---- It points to the address of the space behind the array , I don't know when I will meet '\0', So it's a random value 
    
printf("%d\n", strlen(&arr[0]+1));【5】----strlen The calculation is '\0' The number of previous elements , Yes 5 individual 

char *p = "abcdef";
printf("%d\n", sizeof(p));【4/8】---- because p Is a pointer variable , The size of the pointer variable is 4/8 Bytes , Depends on the compiler 
    
printf("%d\n", sizeof(p+1));【4/8】---- hypothesis p The address for 0x0012ff40 that p+1 Namely 0x12ff41   Or an address , It's the address ,sizeof（ Address ） The size is 4/8 Bytes 
    
printf("%d\n", sizeof(*p));【1】---- because p yes char* Type of , Access a byte , So a byte is a, You get 1
    
printf("%d\n", sizeof(p[0]));【1】----p[0]   <==>  *(p+0)   Is the first character a Size ,a Is the character , So it is 1 Bytes 
    
printf("%d\n", sizeof(&p));【4/8】----&p What is taken out is the pointer variable p The address of , It should be saved with a secondary pointer , But the secondary pointer is also a pointer , So it is 4/8 Bytes 
    
printf("%d\n", sizeof(&p+1));【4/8】----&p+1 Get the address , But the special thing is , What you get is a pointer variable p skip p Variable space , The address of the space behind , It's the address , still 4/8 Bytes 
    
printf("%d\n", sizeof(&p[0]+1));【4/8】----&p[0] yes a The address of ,&p[0]+1 Get is b The address of , It's the address , Namely 4/8 Bytes 
    
printf("%d\n", strlen(p));【6】----p It's in there a The address of , So pass it on to strlen Of course a The address of ,strlen The calculation is '\0' The number of previous elements , Yes 6 individual 
    
printf("%d\n", strlen(p+1));【5】----p It's in there a The address of ,p+1 Get is b The address of , So pass it on to strlen Of course b The address of ,strlen The calculation is '\0' The number of previous elements , Yes 5 individual 
    
printf("%d\n", strlen(*p));【 Compile error 】----*arr Get the first character 'a',strlen（97）, Pass to strlen Should be the address , Dereference to get a in the future ,a The address of is a wild pointer 

printf("%d\n", strlen(p[0]));【 Compile error 】----*arr Get the first character 'a',strlen（97）, Pass to strlen Should be the address , Dereference to get a in the future ,a The address of is a wild pointer 
    
printf("%d\n", strlen(&p));【 Random value 】----&p Get is p The address of , And the string is not a space at all , I don't know when I will meet \0 stop , So it's a random value 
    
printf("%d\n", strlen(&p+1));【 Random value 】----&p+1 Get is p add 1 The address of , And the string is not a space at all , I don't know when I will meet \0 stop , So it's a random value 
    
printf("%d\n", strlen(&p[0]+1));【5】----&p[0] Get is a The address of ,a The address of +1 Get is b The address of ,trlen The calculation is '\0' The number of previous elements , Yes 5 individual 

Two dimensional array

// Two dimensional array 
int a[3][4] = {
    0};
printf("%d\n",sizeof(a));【48】---- The array list is placed alone in sizeof Inside , It calculates the size of the entire array   So it is 48
    
printf("%d\n",sizeof(a[0][0]));【4】----arr[0][0] Is the first column element of the first row of the array , yes int Type of , So it is 4

printf("%d\n",sizeof(a[0]));【16】---- When we access a two-dimensional array ,arr[0][j] j The value range of is 0-3  When accessing a one-dimensional array ,j The value range of is 0-9  Two dimensional array arr[0]  arr[1]   arr[2] Is the array name of each row of the two-dimensional array 
     The array list is placed alone in sizeof Inside , It calculates the size of all elements in that line , So it is 16
     
printf("%d\n",sizeof(a[0]+1));【4/8】----a[0] <==>&a[0][0],&a[0][0]+1 What you get is the address of the second element in the first line , Because it's the address , therefore sizeof（ Address ） The size is 4/8 Bytes 
    
    printf("%d\n",sizeof(*(a[0]+1)));【4】----a[0] <==>&a[0][0],&a[0][0]+1 What you get is the address of the second element in the first line , Because it's the address , Dereference it , What you get is int The size of the type 

printf("%d\n",sizeof(a+1));【】----a Is the array name of a two-dimensional array ,a Represents the address of the first element , The first element of a two-dimensional array is its first row ,a It's the address on the first line , Address of array +1 What you get is the address on the second line , Because it's the address , So it is 4/8 Bytes 
    
printf("%d\n",sizeof(*(a+1)));【16】----*(a+1) <==> a[1]  therefore sizeof(*(a+1)) <==> sizeof(a[1]),a[1] Is the array name in the second line , Calculate the size of the second line 
    

printf("%d\n",sizeof(&a[0]+1));【4/8】----&a[0] I got the address on the first line ,&a[0]+1 What you get is the address of the element in the second line , Because it's the address , So it is 4/8 Bytes 

printf("%d\n",sizeof(*(&a[0]+1)));【4】----&a[0] I got the address on the first line ,&a[0]+1 What you get is the address of the element in the second line , Dereference it , What you get is 4

printf("%d\n",sizeof(*a));【16】----a Is the address of the first element , The first element of a two-dimensional array is the first row , Dereference the address in the first line , What you get is the first line , The first line is placed separately sizeof Inside , What you get is 16

printf("%d\n",sizeof(a[3]));【16】---- Although there is no third line , But it will be analyzed according to the type , obtain 16
 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 .