C中的指针增量

Question

I am getting an odd error in a small C program in which I'm trying to learn how to use pointers.

I assumed everything was working because it compiled but when I try to run the code, it says C program has stopped responding. I must be doing something incorrect... Any help?

#include <stdio.h>


int main()
{

int k = 22;
int z = 11;
int *ptr = k;

 printf("%d \n",*ptr);
 ptr++;
 printf("%d \n",*ptr);




    getchar();
    return 0;
}

Answer 1

You need to assign address of k not it's value..

int *ptr = &k;

Also after ptr++; *ptr would print garbage value(though behavior is undefined)

Answer 2

Your problem is here:

int *ptr = k;

you are assigning the value of k to ptr not the address of, so you are missing the address operator( & ):

int *ptr = &k;
           ^

gcc by default gives the following warning:

warning: initialization makes pointer from integer without a cast [enabled by default]

So if you are not seeing a waring then you should enable warnings.

Performing indirection on ptr after you increment here:

ptr++;
printf("%d \n",*ptr);
               ^
               indirection

it is undefined behavior in this case since it will be pointing to one past the last element, since pointers to non-arrays are treated as 1 element arrays for this case as per C99 section 6.5.6 Additive operators paragraph 7 .

Answer 3

It is very simple to understand pointer, the pointer is a value to store an memory address, and you must always ensure the memory address is avaiable!

so here is the 1st error;

int *ptr = k;

so 1st you shall assign an valid memory address to the ptr before you use it, correct one shall be

int *ptr;
ptr = &K

star (*) is use to get the value which is stored in the memory address you assigned. so to increase the value , you shall

(*ptr)++ 

instead of

ptr++

this will increase the address instead of value itself, and you r really unknown what is stored there.