Unquote the result of a function defined by a macro - macros

Question

I have the following definition of my macro
defmacro defverified(sign, body) do
{name, _, [param]} = sign
quote do
def unquote(name)(unquote(param)) do
unquote(param) = verify! param
unquote(body)
end
end
end
Were verify!/1 returns its paramters if it is verified as a correct parameter
And my function is defined as follows
defverified toto(p) do
IO.inspect p
end
And the inspection of the content of p is correct but the return of the function is a quoted form of my variable.
iex(3)> res = Toto.toto(1)
1
[do: 1]
iex(4)> res
[do: 1]
Is it possible to have an unquoted form for the return of my function or should I unquote it manually?
I expect the following output of my function
iex(3)> res = Toto.toto(1)
1
1
iex(4)> res
1

Answer 1

This happens because the do...end construct is a peculiar piece of syntactic sugar. For example, this:
def toto(p) do
IO.inspect p
end
is equivalent to:
def toto(p), do: IO.inspect p
which, because keywords at the end of an argument get passed as a keyword list, is equivalent to:
def(toto(p), [do: IO.inspect(p)])
Since your defverified macro only expects a do block and no other keyword parameters, we can explicitly match out the actual body:
defmacro defverified(sign, [do: body]) do
...
Before doing this, the last form in the toto function would literally be:
[do: IO.inspect param]
which would call IO.inspect as expected but then return a keyword list instead of a plain value.