Question

Here is a code snippet I was working with:

int *a;
int  p = 10;

*(a+0) = 10;
*(a+1) = 11;
printf("%d\n", a[0]);
printf("%d\n", a[1]);

Now, I expect it to print

10
11

However, a window appears that says program.exe has stopped working.
The if I comment out the second line of code int p = 10; and then tun the code again it works.

Why is this happening? (What I wanted to do was create an array of dynamic size.)

Answer 1

There are probably at least 50 duplicates of this, but finding them may be non-trivial.

Anyway, you’re defining a pointer, but no memory for it to point at. You’re writing to whatever random address the pointer happened to contain at startup, producing undefined behavior.

Also, your code won’t compile, because int *a, int p = 10; isn’t syntactically correct — the comma needs to become a semicolon (or you can get rid of the second int, but I wouldn’t really recommend that).

In C, you probably want to use an array instead of a pointer, unless you need to allocate the space dynamically (oops, rereading, you apparently do want to — so you need to use malloc to allocate the space, like a = malloc(2); — but you also want to check the return value to before you use it — at least in theory, malloc can return a null pointer). In C++, you probably want to use a std::vector instead of an array or pointer (it’ll manage dynamic allocation for you).