Non-greedy regex quantifier is greedy

Question

Note: I am not looking for a working substitution for my command (allready have that), but for an explanation.

I have the following code:

include <stdio.h>

char ga[] = "abcdefghijklm";

void my_array_func(char ca[10]);
void my_pointer_func(char *pa);

int main()
{
   /* code */

   return 0;
}

and use the following search on it:

/^$\n\(^.*\n\)\{-}\ze$\nint main

Which should match:

void my_array_func(char ca[10]);
void my_pointer_func(char *pa);

But it doesn't. It does match all possible lines above. In this case:

char ga[] = "abcdefghijklm";

void my_array_func(char ca[10]);
void my_pointer_func(char *pa);

If I add more lines which are fitting to the pattern it matches them too. So it behaves exactly the same if I replace \{-} with *.

If I use the count directly (\{2} instead of \{-}) I get the result as expected.

Answer 1

This greedyness comes from * that is inside parenthesis. It matches also empty lines. If you change that to \+ it should work as you expect it to.