当前位置：网站首页>Explain sizeof, strlen, pointer, array and other combination questions in detail

Explain sizeof, strlen, pointer, array and other combination questions in detail

2022-07-03 08:21:00 【On the Pearl River】

This part belongs to C Advanced language play

Catalog

This part belongs to pointer 、 Array 、sizeof and strlen Advanced gameplay combined , Help you win easily ！

Summary of pit points

1. sizeof

①sizeof( Array name ); The array name here represents the entire array , Instead of the address of the entire array

②sizeof It's the operator , Calculate the memory space occupied by the type of variable , Unit is byte , Don't pay attention to the contents stored in memory

③sizeof Is based on type analysis , Will not access specific content

④printf("%d\n",sizeof( The pointer )), The size obtained is 4 Bytes （ stay x86 In the environment ） or 8 Bytes （ stay x64 In the environment ）

2. strlen

①strlen It's a library function , When using, you need a header file <string.h>, Specifically for string length , Only for Strings , Keep looking back \0, Statistics \0 The number of characters before

②strlen() You can only receive addresses ,strlen('a') or strlen(97), Illegal access , It's a wild pointer ！

title

The following topics cover Understanding of logarithmic group names 、 The operation of pointer and the meaning of pointer type And other very important knowledge points ,

// One dimensional array

Key summary ：

①sizeof() Only pay attention to the types in brackets , Don't pay attention to its content

②sizeof( The pointer ) The answer is 4 Bytes or 8 Bytes

int a[ ] = {1,2,3,4};
printf("%d\n",sizeof(a)); have to 16
printf("%d\n",sizeof(a+0)); have to 4 or 8
printf("%d\n",sizeof(*a)); have to 4
printf("%d\n",sizeof(a+1)); have to 4 or 8
printf("%d\n",sizeof(a[1])); have to 4
printf("%d\n",sizeof(&a)); have to 4 or 8
printf("%d\n",sizeof(*&a)); have to 16
printf("%d\n",sizeof(&a+1)); have to 4 or 8
printf("%d\n",sizeof(&a[0])); have to 4 or 8
printf("%d\n",sizeof(&a[0]+1)); have to 4 or 8

// A character array

Key summary ：

①strlen() Only address can be stored in brackets , And follow this address all the way back \0

② The character pointer stores the address of the first character , It is not the same address as the address of this pointer

char arr[ ] = {'a','b','c','d','e','f'};
printf("%d\n", sizeof(arr)); have to 6
printf("%d\n", sizeof(arr+0)); have to 4 or 8
printf("%d\n", sizeof(*arr)); have to 1
printf("%d\n", sizeof(arr[1])); have to 1
printf("%d\n", sizeof(&arr)); have to 4 or 8
printf("%d\n", sizeof(&arr+1)); have to 4 or 8
printf("%d\n", sizeof(&arr[0]+1)); have to 4 or 8
printf("%d\n", strlen(arr)); Get a random value
printf("%d\n", strlen(arr+0)); Get a random value
printf("%d\n", strlen(*arr)); have to error
printf("%d\n", strlen(arr[1])); have to error
printf("%d\n", strlen(&arr)); Get a random value
printf("%d\n", strlen(&arr+1)); Get a random value -6
printf("%d\n", strlen(&arr[0]+1)); Get a random value -1

char arr[ ] = "abcdef";
printf("%d\n", sizeof(arr)); have to 7
printf("%d\n", sizeof(arr+0)); have to 4 or 8
printf("%d\n", sizeof(*arr)); have to 1
printf("%d\n", sizeof(arr[1])); have to 1
printf("%d\n", sizeof(&arr)); have to 4 or 8
printf("%d\n", sizeof(&arr+1)); have to 4 or 8
printf("%d\n", sizeof(&arr[0]+1)); have to 4 or 8
printf("%d\n", strlen(arr)); have to 6
printf("%d\n", strlen(arr+0)); have to 6
printf("%d\n", strlen(*arr)); have to error
printf("%d\n", strlen(arr[1])); have to error
printf("%d\n", strlen(&arr)); have to 6
printf("%d\n", strlen(&arr+1)); Get a random value
printf("%d\n", strlen(&arr[0]+1)); have to 5

char *p = "abcdef";
printf("%d\n", sizeof(p)); have to 4 or 8
printf("%d\n", sizeof(p+1)); have to 4 or 8
printf("%d\n", sizeof(*p)); have to 1
printf("%d\n", sizeof(p[0])); have to 1
printf("%d\n", sizeof(&p)); have to 4 or 8
printf("%d\n", sizeof(&p+1)); have to 4 or 8
printf("%d\n", sizeof(&p[0]+1)); have to 4 or 8
printf("%d\n", strlen(p)); have to 6
printf("%d\n", strlen(p+1)); have to 5
printf("%d\n", strlen(*p)); have to error
printf("%d\n", strlen(p[0])); have to error
printf("%d\n", strlen(&p)); Get a random value
printf("%d\n", strlen(&p+1)); Get a random value
printf("%d\n", strlen(&p[0]+1)); have to 5

// Two dimensional array

Key summary ：

① A two-dimensional array arr[3][4],arr[0] Represents the address of the first line ,arr[1] Indicates the address of the second line ,arr[2] Indicates the address of the third line

②sizeof('a'+1), Integer promotion will occur

int a[3][4] = {0};
printf("%d\n",sizeof(a)); have to 48
printf("%d\n",sizeof(a[0][0])); have to 4
printf("%d\n",sizeof(a[0])); have to 16
printf("%d\n",sizeof(a[0]+1)); have to 4 or 8
Particular attention ： In this question a[0] Represents the address of the first element in the first line ,+1 Is the address of the second element in the first line
printf("%d\n",sizeof(*(a[0]+1))); have to 4
printf("%d\n",sizeof(a+1)); have to 4 or 8
Particular attention ： In this question a Represents the address of the first row of the two-dimensional array ,+1 Is the address of the second row of the two-dimensional array
printf("%d\n",sizeof(*(a+1))); have to 16
printf("%d\n",sizeof(&a[0]+1)); have to 4 or 8
Particular attention ： In this question &a[0] Represents the address of the first row of the two-dimensional array ,+1 Is the address of the second row of the two-dimensional array
printf("%d\n",sizeof(*(&a[0]+1))); have to 16
printf("%d\n",sizeof(*a)); have to 16
Particular attention ： In this question a Represents the address of the first line of elements , The first row of the two-dimensional array after dereference
printf("%d\n",sizeof(a[3])); have to 16
Particular attention ： Even though a[3] It belongs to cross-border visit , however sizeof Calculate the memory space occupied by the type of variable ,sizeof Analysis by type , The fourth line will not be accessed