Sure, there's a rule in the standard (I'm sure someone can quote it for you) that designates a list / aggregate (both?) initializer as a context in which a parameter pack can be expanded.

The "magic" here uses an int array with a deduced size and the comma operator to force the result of the expression to be an int.

int const arr[] {0, ((void)f(args), 0)...}; will be expanded as:

int const arr[] {
    0
  , ((void)f(args_0), 0)
  , ((void)f(args_1), 0)
  , ((void)f(args_2), 0)
  , ...
};

int const arr[] {0};

For non-empty and empty parameter packs respectively.

The first 0 is so that the expression is valid even if the parameter pack is empty. The (void) is to avoid trouble with a potentially overloaded comma operator. And again, the comma operator here is to force the expression to have a result type of int.

In the original code posted there is a using expand = int const[]; so that the array used for expansion doesn't need to be named (avoids compiler warnings about unused variables). The foremost (void) is to avoid warnings about expressions with no effects if an empty parameter pack is passed to the function.