Preliminary cognition of C language pointer

1 How variables are accessed

2 The definition of a pointer variable

3 actual combat

1 Two ways to access variables ：1 By variable name 2 adopt Variable address Let's give you an example  

such as xx The flower shop Where are you going You need to know the name of the florist perhaps yes The address of the florist such You can get to .

Two ways to access variables Code It's a demonstration

#include<stdio.h>
void main(void)
{
   int a=10;
    printf("%d\n",a);// Access... By variable name 
    printf("%d\n",*(&a));// Access by address   

}

#include<stdio.h> Header file means Include input and output functions in a specific path

Direct use a Namely Variable name visit perhaps Take the address of the variable name stay * Is to take out the value of the address These are two ways to access variables The end result is the same .

2 Pointer variable meaning ： A variable that holds the address The pointer Address again So simply understand The pointer == Address Address == The pointer

The definition of a pointer variable

int *p;

Define a int Pointer variable of type * Is an ID And tell the computer to declare that I am a pointer variable

Application of pointer variables

#include<stdio.h>

void main(void)
{
 int a=10; // Define a int  Type of a Variable 
 int *p;// Define a pointer variable   * Is an identifier.    Tell the computer to declare that I am a pointer variable 
  p=&a;// initialization 

printf("  The value obtained by accessing the address  ：%d",*p);// How to write it 1
printf(" It is also the value obtained through address access ：%d",*(&a));// And writing 1  equally 
printf(" Value obtained by variable name  %d",a);
  

}

Pointer variables need Type required   We illustrate with several examples  

First Let's define a char Pointer to type Variable and int Variable of type size use sizeof keyword Print out   

See how it turns out  

#include<stdio.h>

void main(void)
{
    int *p=10;
    char *a=2;

     printf("int   Pointer variable size of type  %d\n",sizeof(*p));
     printf("char  Pointer variable size of type  %d",sizeof(*a));

}

    Defined a int Pointer variable of type   It's a point int Variable of type   char That's the same understanding  

So in the printout The result is 4 and 1 

int The type is 4 Bytes   char yes A byte   

We are adopting a method   Come on prove Pointer to the variable need Claim type statement

Print by offset Address contrast  

#include<stdio.h>

void main(void)
{
    int  *p;
    char *a;

    printf("int   No offset   %#x\n",p);
    printf("int   With offset   %#x\n",++p);
    printf("\n");
    printf("char  type   No offset   %#x\n",a);
    printf("char   The type has an offset   %#x\n",++a);
 
   

}

Running results  

  Why does this result  

1 int The type is Four byte size     Self adding offset One is Four bytes

2 char The type is 1 Byte size   Self adding The offset One is A byte  

The problem of address offset Is it right? The offset of the whole block , Not at all  

First use int Examples of types   It has a size of four bytes   We In operation offset when   It is active within its four byte range . Another way int type It's four bytes   Offset by one   （++p）int The type becomes 8 Bytes Obviously it's not right   . It's like A vernier caliper   It's so big You can move The offset   To different addresses  

in general   Pointer variables require defined types       because It determines 1 Point to the size of the space 2 Determines the offset  

int The type remember is Four byte size   char The type is a byte size