C Programming - Different signature problem with extern variables

One of my friend asked me this question few days back which seemed to be very interesting for me atleast. The question was as follows

File1.c		File2.c
1 int arr[80];		1 extern int *arr; 2 int main() 3 { 4 arr[1] = 100; 5 return 0; 6 }

What will be the output when complied and linked together?

It was surprising for me to see a segmentation fault for this kind of a code. Even though array and pointers are different in aspects like
1. Array variable is const pointer
2. Array size has to be defined while declaring the variable

Still I thought array will be manipulated like a pointer, but here in this case what is happening is when we are putting the value 100 to arr[1] it taking the value of the pointer here in this case 0 as arr[80] will be filled with zeros and then adding one to that which mean resulting value will be 0x4 as it is a pointer addition. After that it is trying to put 100 into the address pointed by 0x4 which results in a segmentation fault.

I don't have a sincere justification for this other than explaining why it is working like this. If anyone have any response please post it as a comment.

Labels: C

About this entry

You’re currently reading “C Programming - Different signature problem with extern variables,” an entry on ANOOB BACKER Technical Blog

Published:: 8:18 am; by -

Archives

Pramod 9:17 am

(1) int arr[80];
(2) int *arr;

are different things. (1) will "create space" for 80 integers in the data segment. "arr" is just shorthand for that memory location (in the data segment). (2) will "create space" for one pointer in the data segment.

So when you say:

extern int* arr;

You are actually telling the compiler to assume there's exactly one pointer in some other translation unit. When you do arr[1]=100, the code thats going to be generated in x86 assembly would be something like.

mov esi, arr
mov [esi+4], 100

key point here is that we are moving the "contents of arr" into esi.

in the case of an array, the code would look exactly the same, with one major semantic difference

mov esi, $arr,
mov [esi+4], 100

here "arr" is a constant (referring to the location where the array begins).

Hence the difference.

Anonymous 6:33 am

I think problem is because of File Scope.
ie int arr[80] will decay to
int *const arr meaning arr is constant pointer to integer.
and constants have file scope.

So question comes why code compiles ?
Answer is code gets compiled in units called Translation units mapping to file and order of which is unspecified.

Anonymous 6:39 am

I think int arr[80] will decay to
int *const arr; during rutime
meaning arr is constant pointer to integer.
Constants have file scope.
So question is why did code compile ?
Answer is code gets compiled in terms of translation units mapping to file order of which is unspecified.

ANOOB BACKER Technical Blog

C Programming - Different signature problem with extern variables

About this entry

3 Comments (Post a Comment)

Recently

Archives

Links