创建一个变量，然后将其内存地址增加1

Question

I created a variable and then I printed its location number. I then incremented the memory location by 1. I got the desired results in two cases while one case is giving an answer which could not be explained properly by my teachers. It would be really helpful if somebody told me why the second printf("%p\\n",&i) is giving the same output that was given first time.

int main()
{
    int i=3,*x;
    x=&i;
    printf("%p\n",x);
    printf("%p\n",&i);
    printf("%p\n",&(*x));
    x++;
    printf("\n");
    printf("%p\n",x);
    printf("%p\n",&i);
    printf("%p\n",&(*x));
}



0x7ffce5dc5208
0x7ffce5dc5208
0x7ffce5dc5208

0x7ffce5dc520c
0x7ffce5dc5208
0x7ffce5dc520c

Answer 1

Every object in a C program has a unique fixed address during its existence. You store a copy of i 's address into x . Then you manipulate x to change the value it holds, it now contains another address. But the manipulation of the value in x does not move i . The object named i is still in the same spot.

Answer 2

The address of a variable never changes during its lifetime. Just because x starts out containing the value of that address and you then increment the value of x doesn't mean that the address of i changes.

Taking an analogy from the comments, suppose you write the address of your house on a piece of paper. Now suppose you increment the value of the house number on that sheet of paper. Your home is not in a different place, you just changed the house number you wrote down.

So &i will always have the same value, regardless of whatever value x may contain.

Answer 3

Because the pointer value is incremented but the address of the variable i remain same. Pointer variable is also a variable that holds the address of variable i , changing its value won't effect the address of i variable.

Answer 4

The piece of code x = &i copies i 's address into the x pointer. It's a copy, so modifying it only modifies the value stored in x , never will it change i 's address.

I'm not sure what you are trying to achieve though. You can't change the address of a variable.

Answer 5

This has nothing to do with pointers and memory address, specifically. Check the below program

int x = 5;
int y = x;

printf (" x= %d, y = %d\n", x, y);
y++;
printf (" x= %d, y = %d\n", x, y);

Output:

x= 5, y = 5

x= 5, y = 6

Changing y does not affect the value of x . Same is applicable in your case, too.

Simply put, x and i are two different variables, they have their own values. Now, given the fact that x is a pointer and assigned with the value of the address of i , you can use x to get the value stored at i . But that's it.

This does not somehow "link" the two variables together, they still can be modified independent of each other. Only thing, after modifying the value of x , you can no longer use that to access the value stored in i , because it does not point to the address of i anymore.