Question

EDIT: Originally I had transcribed i++ not i--

The code now is as it was, and the code in the code block compiles and works.

Why, if unsigned int i; is used instead of int i; in the code snippet below, does using the function result in a segfault?

void insertion_sort_int_array(int * const Ints, unsigned int const len) {
     unsigned int pos;
     int key;
     int i;

     for (pos = 1; pos < len; ++pos) {
         key = Ints[pos];
         for (i = (pos - 1); (i >= 0) && Ints[i] > key; i--) {
             Ints[i + 1] = Ints[i];
         }
         Ints[i + 1] = key;
     }
}

Answer 1

insertionSort(array A)
begin
    for x := 1 to length[A]-1 do
    begin
        value := A[x];
        i := x - 1;
        while i >= 0 and A[i] > value do
        begin
            A[i + 1] := A[i];
            i := i - 1;
        end;
        A[i + 1] := value;
    end;
end;

The only difference between the standard insertion sort algorithm and your code is that you’re incrementing i instead of decrementing. That’s your problem. I bet that in the code you’re actually compiling and running, you have i– instead of i++ in the inner loop. That’s why the unsigned i makes a difference – it cannot be negative, so the inner loop will never end. Did you copy the code wrong when you posted?

EDIT:

Well, now that you changed the posted code, it all makes sense, right? An unsigned i will simply underflow to INT_MAX when you decrement it past 0, which will cause you to access memory outside of the array bounds.