Initializer list in user-defined literal parameter

Question

I don\'t know if it\'s possible but I want to do stuff like

int someval = 1;
if({1,2,3,4}_v.contains(someval ))

but when I try to define l

Answer 1

§13.5.8/3 says:

The declaration of a literal operator shall have a parameter-declaration-clause equivalent to one of the following:

const char*
unsigned long long int
long double
char
wchar_t
char16_t
char32_t
const char*, std::size_t
const wchar_t*, std::size_t
const char16_t*, std::size_t
const char32_t*, std::size_t

So it looks like you can't have a parameter of initializer_list type.

I can only think of the obvious as an alternative; if you don't mind typing a little more you can do something like

std::vector v(std::initializer_list l) {
    return { l };
}

int someval = 1;
if(v({1,2,3,4}).contains(someval))

Alternatively you could get wacky and write an operator overload for initializer_list (haven't tested though):

bool operator<=(std::intializer_list l, int value) {
    return std::find(std::begin(l), std::end(l), value) != std::end(l);
}

And

if ({1, 2, 3, 4} <= 3)

should work...

Actually nevermind, it doesn't. You'll have to go with a normal function.