![]() |
Home | Libraries | People | FAQ | More |
Return a reference to the underlying std::int64_t,
or throw an exception.
std::int64_t& as_int64();
If is_int64()
is true, returns a reference
to the underlying std::int64_t,
otherwise throws an exception.
Constant.
Strong guarantee.
|
Type |
Thrown On |
|---|---|
boost::system::system_error |
|
This function is intended for direct access to the underlying object,
if it has the type std::int64_t.
It does not convert the underlying object to type std::int64_t
even if a lossless conversion is possible. If you are not sure which
kind your value has,
and you only care about getting a std::int64_t
number, consider using to_number instead.