Issue with char and string overlap

[JS]

I'm trying to create a split function that can handle both char 
and string delims. I initially created two separate functions but 
this doesn't work for default parameters since the compiler 
doesn't know which one to choose(but in this case both would work 
fine and it would be nice to inform the compiler of that). I then 
tried to template and conditionally code the two but it still 
doesn't work:

both functions work separately but when i uncomment the string 
version I get an error about the string version shadowing.

import std.stdio, std.cstream;

string[] split(T)(string s, T d) if (is(T == char) || is(T == 
string))
{
	int i = 0, oldj = 0; bool ok = true;
	string[] r;
	foreach(j, c; s)
	{
		static if (is(T == char))
		{
			if (c == d)
			{
				if (!ok) { oldj++; continue; }
				if (r.length <= i) r.length += 5;
				r[i] = s[oldj..j];
				i++; oldj = j+1;
				ok = false;
			} else if (!ok) ok = true;
		}
		else if (is(T == string))
		{
		/*
			for(int j = 0; j < s.length - d.length; j++)
			{
				if (s[j..j + d.length] == d)
				{
					if (!ok) { oldj++; continue; }
					if (i == r.length) r.length += 5;
					r[i] = s[oldj..j - d.length + 1];
					i++; oldj = j + d.length;
					ok = false;
				} else if (!ok) ok = true;
		
			}
		*/
		}
	}
	if (oldj < s.length)
	{
		if (r.length <= i) r.length++;
		r[i] = s[oldj..$];
		i++;
	}
	r.length = i;
	return r;
}


string[] splitS(string s, string d = " ")
{
	int i = 0, oldj = 0; bool ok = true;
	string[] r;
	for(int j = 0; j < s.length - d.length; j++)
	{
		if (s[j..j + d.length] == d)
		{
			if (!ok) { oldj++; continue; }
                         if (r.length <= i) r.length += 5;
			r[i] = s[oldj..j - d.length + 1];
			i++; oldj = j + d.length;
			ok = false;
		} else if (!ok) ok = true;
		
	}
	if (oldj < s.length)
	{
		if (r.length <= i) r.length++;
		r[i] = s[oldj..$];
		i++;
	}

	r.length = i;
	return r;
}

void main(string[] args)
{

	auto s = splitS("abc bas   ccc", " ");
	
	foreach(a; s) writeln(a);
	
	din.getc();
}