strcpy?

[CornedBee]

Why the -1 jim? Isn't it confusing if your function uses a 1-based index?
And for the sake of good style the source should be const char*.
The code is quite simple.
The arguments are:
dest: the destination. A char buffer at least 2 chars long.
src: the source string.
where: the 1-based index into src of the character to copy.

*dest=*(src + (where -1) );
*dest
dest is dereferenced. This means the assign operation assigns to the character dest points at, not the pointer itself. It is equivalent to
dest[0]

*(src + (where -1) )
This is equivalent to
arc[where-1]
The -1 is because where is 1-based but arrays in C are 0-based.
So this takes the whereth character of src and copies it to the first position in dest.

*(dest+1)=0x00;
This adds a terminating NUL-character to dest in order to make it a valid C-string.
*(dest+1)
is equivalent to
dest[1]
so it accesses the second position of the buffer (the first is where the copied character is).

return;
This line is redundant in ANSI C but probably required by some of the weird compilers jim has or had to work with.