Question

In kernel.h min is defined as:

#define min(x, y) ({                \
    typeof(x) _min1 = (x);          \
    typeof(y) _min2 = (y);          \
    (void) (&_min1 == &_min2);      \
    _min1 < _min2 ? _min1 : _min2; })

I don’t understand what the line (void) (&_min1 == &_min2); does. Is it some kind of type checking or something?

Answer 1

The statement

(void) (&_min1 == &_min2);

is a guaranteed “no-op”. So the only reason it’s there is for its side effects.

But the statement has no side effects!

However: it forces the compiler to issue a diagnostic when the types of x and y are not compatible.
Note that testing with _min1 == _min2 would implicitly convert one of the values to the other type.

So, that is what it does. It validates, at compile time, that the types of x and y are compatible.