試圖了解in指針的引用和地址

Question

給定下面的代碼，為什么a和b的值會發生變化。.P1和p2存儲a和b的地址，為什么當p1和p2發生變化時a和b也會發生變化

#include <stdio.h>


int main(int argc, char** argv) 
{

//&a stores address
//*p1 is a pointer

int a = 5, b = 10;
 int *p1, *p2;  //p1 and p2 are 2 pointers which can store int addressses

 p1 = &a;  //now p1 and p2 stores address of a and b
 p2 = &b; 



  printf("p1 storing address of a  = %d\n", *p1); 
 printf("p2 storing address of  b = %d\n", *p2);

 *p1=30;  
 *p2=40;

 printf("p1 assigning values to p1 pointer = %d\n", *p1); 
 printf("p2 assigning values to p2 pointer= %d\n", *p2);



 printf("a whose value is = %d\n", a); 
 printf("b = %d\n", b);

}

Answer 1

我認為，如果您還自己打印指針而不是僅打印指針的內容，那么事情對您來說會更加清楚。 您可以使用%p說明符。

#include <stdio.h>

int main()
{
  int a = 10;
  int b = 20;
  int *p = &a;
  int *q = &b;

  printf("1) p is %p\n", p);
  printf("1) q is %p\n", q);
  printf("1) a is %d\n", a);
  printf("1) b is %d\n", b);

  p = q;

  /* At this point, I changed the value of p so that it  points to `b`
     just like `q` does. `a` and `b` are still unchanged. */ 
  printf("2) p is %p\n", p);
  printf("2) q is %p\n", q);
  printf("2) a is %d\n", a);
  printf("2) b is %d\n", b);

  *p = 30;

  /* Now that p points to `b`, dereferencing the pointer will affect `b`
     instead of `a` */
  printf("3) p is %p\n", p);
  printf("3) q is %p\n", q);
  printf("3) a is %d\n", a);
  printf("3) b is %d\n", b);
}

當您說*p = something您正在分配p指向的內存位置（根據設置方式的不同，可能是a或b ）。 另一方面，如果執行p = q ，則將更改指針本身，而不是指針所指向的內容。

Answer 2

p1和p2包含地址a和b 。 當我們使用*p1=30; 它告訴編譯器將p1包含其地址的變量的值分配為30

您可以將其視為由p1表示的地址值為30 。

* = value at

和

& = address of