指针地址和指针包含的地址之间的差异

Question

#include<stdio.h> 

main() {
  int a,*p;
  a=2;
  p = &a;

  printf("The address %p contains the address of a=2:%p\n",&p,p);

  printf("The address of %d is %p\n",a,&a);

  return 0;
}

The result is:

The address 0028FF18 contains the address of a=2: 0028FF1C

is this correct? if not, what's the correct statement??

The address of 2 is 0028FF1C

is this correct? if not, what's the correct statement?? */

Answer 1

p有其自己的地址＆p，但p的值是a的地址。

Answer 2

Regarding a:

printf("The value of a is %d.", a); // The value of a is 2.
printf("The address of a is %p.", &a); // The address of a is 0028FF1C

Regarding p:

printf("The value of p is the address of a is %p.", p); // The value of p is the address of a is 0028FF1C
printf("The address of p is %p.", &p); // The address of p is 0028FF18

I think of memory on a computer as a street full of addresses (a really long street), and at each address lives a single value. The variables a and p are simply syntax needed for us as programmers to conceive of these memory locations. So, a is a representation for the address 0028FF1C and the current resident of that address is 2 ... in the future, 2 could move out and 4 could move in, but the address would be the same.

Likewise with p - his address will always be 0028FF18, and it just so happens that the address of a is the current resident at this location, but later a different value could move in here (eg, p = &b ) and p 's address would still be 0028FF18.

Answer 3

Look at the diagram below:

+---------+
|         |
|    2    | Value of a=2
|         |
+---------+
Address=0028FF1C


+---------+
|         |
|0028FF1C |  P stores address of a
|         |  So p=0028FF1C
+---------+
Address=0028FF18

So now your first question is

1. The address 0028FF18 contains the address of a=2: 0028FF1C. Is this correct?

Answer: Yes, it is correct as 0028FF18 is address of p and it contains &a ie address of a which is 0028FF1C.

Now coming to second question

2. The address of 2 is 0028FF1C. Is this correct?

Answer: This is little bit confusing statement (can't say it's perfectly correct) as address of a is 0028FF1C and value of a is 2. So it might be better if you print it like:

printf("The address of a is %p\n",&a);

So it will give output:

The address of a is 0028FF1C.

This statement seems better as here we are printing address of variable not the address of a value which that variable stores.

Answer 4

A pointer is really just a special kind of number — an address used to locate another piece of data. In your example, p is the numeric address to a , and &p is a pointer to the address stored in p . &p is a pointer to a pointer.

Answer 5

A picture is worth a thousand words, here is what p and a like in memory.

left are address, and right are variable names.

            +---------------+      
  0028FF18  |   0028FF1C    |  p               
            +---------------+                  
  0028FF1C  |      2        |  a                
            +---------------+      

Answer 6

In your situation, both are correct!

How to understand pointer easily?

Just imagine a variable as a box with label(variable name) on it, and the box itself has an address(memory address) ,such as Room 304 on 3rd floor, for anyone to locate it. In the box lays a note(value) tells you what it is(a int 3 , a char 's' , a float 3.14 etc).

While a pointer is also a box with the exception that it contains a note with another box's address on it.

How you read a int ?

How it works on an variable int a=2 ?

You just say, hey I want to access a (the box with label a on it), someone(your compiler along with OS) just find the box and give it to you. You open the box, and see a note reading 2 (might as well as type int ). You can change it with another note you just scribbled, say, 3 .

How you read a int * ?

Similarly, say we have int *p = a . We get the box with label p on it, open it, and it says Room 305 on 5th floor . So instead of saying 'I want the box a ', this time we say 'bring the box in Room 305 on 5th floor'(which is what *p means).

• *p = 3; means: bring the box which box p points to and change its value to 3.
• int *q = &a : bring a new box(put it somewhere), and fill it with a note reads the address of a
• &p : give me the address of box p

Conclusion

1. Every variable has a memory address, and a value
2. A normal variable just deals with its value directly
3. A pointer has a value points to another address(most time, the address of another variable).
4. The above Box Model also works on int **p (pointer to pointer) and more.

Answer 7

Maybe a simple analogy will help you understand better.

int a = 5;
int *p = &a;

• 5 is the value of int variable a .
• &a is the value of int * variable p .
• *p is the int value at the address held by int * variable p .