取消引用指针以键入下一个值时出错

Question

This code should have brought me a printed value of 7. As i am have declared integer i=5 which allocates 4 bytes in memory and integer myInt which also allocates 4 bytes next to the int i . Then i declared a pointer to store the location/address of the int i . But when I add 1 integer to the pointer or i's location which means 4bytes to pointer location of i and then dereference it. I don't get the 7 which i stored in myInt rather i get some value as 243423(something like this)

#include <stdio.h>
#include <stdlib.h>

void main()
{
int i=5;
int myInt=7;
int *pointer = &i;

printf("%i\n", *(pointer+1));
getch();
}

Answer 1

pointer is a different address location pointing to i , now when you add +1 a sizeof(int) it addresses next location of pointer but not second stack variable myInt .

 pointer
+-------+               +------+
| 0x100 | -----> 0x100  |  5   |
+-------+               +------+
  0x300          0x104  |  7   |
                        +------+

A simple memory can be as above, both local variables i & myInt can store in stack in contiguous memory locations. But you made pointer to point to i and pointer itself stored in address location 0x300 . Now when you dereference next memory location of pointer which doesn't point to myInt rather some other memory location which is Undefined Behavior.

Answer 2

Hmm, memory allocation isnt always continuum. *(pointer+1) points to uninitialized memory.

Answer 3

I can see where you're going, but you're getting tripped up by a misunderstanding.

If you allocate an array, the elements of the array will be placed next to each other, and things will work the way you're expecting:

int foo[] = { 4, 3, 2, 1 };
int* i = foo;
printf("%d\n", *i);
printf("%d\n", *(i + 1));

Likewise, if you malloc a block of memory, and then put objects next to each other in that memory, you can use pointer arithmetic the way you're expecting.

However, if you allocate two objects at about the same time, the language does not promise to put them next to each other in memory. There is no reason for pointer arithmetic to work unless you tell the compiler "I want to allocate these objects next to each other." You do that by creating an array, or allocating a block of memory.

---

A compiler is allowed to lay out memory as described in the video, but it's not required to. Using the variables from the video and the original question, the compiler is allowed to notice that you never do anything with myInt and optimize it away. It's also allowed to notice that myInt never changes and output a constant which can be allocated in a different memory area. It might decide to allocate all local variables contiguously on the stack, but in a different order than what appears in the source code (perhaps doing so allows for better use of RAM).

Even in cases where it puts i and myInt next to each other, on some platforms i will have the lower address and on other platforms myInt will. The compiler has enough flexibility that if you want the variables allocated next to each other, you need to make that clear.

Answer 4

Declare "i" and and "myInt" as static and you will get your expected behaviour.

Note 1: This kind of design is only useful for didactic purposes;
Note 2: The correct way to store these values in order to access using pointers is to use an vector.