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Evaluation work on indexing and incrementing

#include <stdio.h>

int main() {
    int op9=9;
    int op99=99;
    int op999=999;
    int op9999=9999;

    int *ptr1=&op9,*ptr2=&op99,*ptr3=&op999,*ptr4=&op9999;

    *ptr1++;
    ++*ptr2;
    (*ptr3)++;
    *++ptr4;

    printf("%d=ptr1 \t %d=ptr2 \t %d=ptr3 \t %d=ptr4",ptr1,ptr2,ptr3,ptr4);

    printf("\n %d = *ptr++ \n %d = ++*ptr \n %d = (*ptr)++ \n %d = *++ptr",*ptr1,*ptr2,*ptr3,*ptr4);

    printf("\n \n  %d=ptr1 \t %d=ptr2 \t %d=ptr3 \t %d=ptr4",ptr1,ptr2,ptr3,ptr4);

    return 0;
}

This is the result I am getting

-

    209002336=ptr1   -209002344=ptr2     -209002348=ptr3     -209002348=ptr4

-209002348 = *ptr++ 
     100 = ++*ptr 
     1000 = (*ptr)++ 
     1000 = *++ptr4

      -209002336=ptr1    -209002344=ptr2     -209002348=ptr3     -209002348=ptr4

Why * ptr ++ which is the point of ptr1 for the address ptr3? and why is * ++ ptr4 value 1000 which is ptr3 value?

I thought * ptr1 ++ would point to the next address, and since it's not part of my program, I expected the program to crash. Likewise * ++ ptr4 should also do the same because it evaluates to * (++ ptr4).

So how does the Increment and Indirection Operator work for * ptr ++ and * ++ ptr?

+3


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3 answers


Explanation 1000 = *++ptr4

:

Formally, this behavior is undefined.

In practice, local variables are allocated on the stack sequentially and in reverse order:

int op9;    // SP+3
int op99;   // SP+2
int op999;  // SP+1
int op9999; // SP+0


So, taking a pointer to op9999

and incrementing it, it ends with op999

.

Keep in mind that this is still undefined behavior according to the C language standard.

This effectively means that you can get different results using different compilers.

+2


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You asked:

Why *ptr++

, which is ptr1

, points to an ptr3

Address?

First of all,

*ptr++;

coincides with

 *(ptr++)


A side effect of the expression is that it is ptr

incremented, the pointer is dereferenced, and the dereferenced value is discarded. The highlight ptr

after it has grown is itself the cause of undefined behavior. Anything you get after that is potentially unreliable.

As for why it ptr

points to an address ptr3

, I'm not sure why you think so. Your conclusion does not seem to indicate this.

You also asked:

and why *++ptr4

does 1000 matter ptr3

?

The expression *++ptr4

suffers from the same problem. You are incrementing ptr4

and dereferencing an incrementing pointer. What you get from dereferencing this pointer causes undefined behavior and therefore unpredictable.

Depending on how your stack is organized, it is quite possible that it ptr4

points to the same address as ptr3

after increasing it. However, do not expect this overlapping behavior.

+1


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The relative memory order int

in your code is undefined. Use an array to place them next to each other in ascending order:

int op[] = {9, 99, 999, 9999};
int *ptr1 = &op[0], *ptr2 = &op[1], *ptr3 = &op[2], *ptr4 = &op[3];

To print pointers, use the format specifier %p

.

0


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