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In order to solve this problem, this article introduces the corresponding analysis and solution in detail, hoping to help more partners who want to solve this problem to find a more simple and feasible method.

Pointer variable and address

Who is the variable for?

A variable is an abstract naming of a space.

A variable name is an alias for a piece of space that you abstract.

The pointer is the address. Point to an address.

Pointer and pointer variable

The pointer points to a block address. The pointer (address) is constant.

Pointer variables can be changed.

# include int main () {int I = 1; int * p = & I; printf ("I =% d\ n", I); printf ("& I =% p\ n", & I); printf ("p =% p\ n", p); printf ("& p =% p\ n", & p); printf ("* p =% d\ n", * p); / / Why not char* p = & I / / TYPE NAME = VALUE / / int* p = & I; / / int I = 1;}

Direct access indirect access

Take up memory space

All 8 bytes, in linux 64-bit.

Null pointer wild pointer null type

Int * I = NULL; pointer operation

Two pointers point to an array at the same time. + +, -, comparison, relationship, &, *

Pointer and one-dimensional array

What is the difference between an array name and a pointer?

An is an array name is a constant that represents an address.

The pointer is a variable.

Astatine +

Pendant +

# include int main () {int a [3] = {1pm 2pm 3}; int * p = a; int i; for (I = 0polii

< sizeof(a)/sizeof(*a); i++) { printf("%d %d %d %d \n",a[i],*(a+i),p[i],*(p+i)); // a[i] printf("%p %p %p %p \n",a+i, &a[i],p+i, p+i); // &a[i] } printf("\n");} 这里代码体现什么是指针常量什么是指针变量？ #include int main(){ int a[3]; int i; int *p = a; for(i = 0;i < sizeof(a)/sizeof(*a); i++) { printf("%p ->

% d\ n ", & a [I], a [I]);} for (I = 0 type name;-> int [3] * ptincincint main () {/ / Array pointer int a [2] [3] = {1m 2M 3M 5J 9}; int iJ; int * p = * a; int (* Q) [3] = a; / / printf ("% d\ n ", * a) Address of / / a [0] [0] / printf ("% d\ n", * * a); / / 1 # if 0 / / printf ("% d\ n", * p); / / Q / / int * p = * a; / / printf ("% d\ n", * p); / / Q / / int (* Q) [3] = aq1; / / printf ("% d\ n", * Q) / / 4 printf ("\ n"); for (I = 0

< 2; i++) { for(j = 0; j < 3; j++) { // printf("%p->

% d\ n ", * (aplasi) + j endif * (* (aqumi) + j)); printf ("% p->% d\ n ", * (Qimi) + j));} printf ("\ n ");} # endif}

Pointer array:

# include # include # include / * int * arr [3];-> TYPE NAME;-> int * [3] arr;*/int main () {char * name [5] = {"english", "math", "cpp", "teacher", "computer"}; int iJournal j; for (I = 0; I

< 5; i++) { puts(name[i]); } for(i = 0; i < 5 ;i++) { int k = i; for(j = i+1;j < 5; j++) { if(strcmp(name[k],name[j]) >

0) {k = j;} if (k! = I) {char * tmp = name [I]; name [I] = name [k]; name [k] = tmp; printf ("sorted:\ n"); exit (0);}

Pointers and functions

Function:

Echo $? / / displays the return value of the previous command

# include using namespace std;/* run this program using the console pauser or add your own getch, system ("pause") or input loop * / / * Definitions: int a [N] = {1meme 2meme 3rem 4m 5je 6}; int * p = a -> a * a [0] & a [3] p [I] p * p pendant 1-> int* int int int* * / void func1 (int* a line int n) void func1 (int* a line int n line int * b) {cout chinese);} return 0;}

Memory alignment problem

Addr/sizeof ()

Structural type-structure internal storage problem and function parameter transfer

Pave the way for the advanced linux in the back.

A student management system can be implemented.

Production and significance

Type description

Nested definition

Define variable

Occupied memory size

Function parameter transfer

Bit domain

Union name {data type member name 1; data type member name 2;}

Enumerated type

Enum name {member 1; member 2; member 3;} what is C language? C language is a process-oriented, abstract general programming language, which is widely used in low-level development. Using C language, low-level memory can be compiled and processed in a simple way.

This is the answer to the question about how to analyze the pointer, function, structure and common body. I hope the above content can be of some help to you. If you still have a lot of doubts to be solved, you can follow the industry information channel for more related knowledge.

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