Question

I am going through C and came to following example which I could not understand from a book “Pointers on C”. Here is the code.

The following declarations are present in one source file:

int     a[10];
int     *b = a;

But in a different source file, this code is written:

extern         int    *a;
extern         int    b[];
int            x, y;
...
x = a[3];
y = b[3];

Can someone explain what happens when the two assignment statements are executed? (Assume
that integers and pointers both occupy four bytes.)

When I tried running the code, it gave me segmentation fault for x and when I comment it out and print the value of y, it gave me 0. The concept was some difference between the pointers and arrays.

Answer 1

What happens is undefined behavior, which in your case manifests itself as a segmentation fault.

You most likely want this instead:

int     a[10];
int     *b = a;

extern         int    *b;
extern         int    a[];

And then both of your assignments will fetch a‘s element at index 3 and put it into x and y.