3 ms·
You can also get the "correct" type (type of &x[0]) by using: *(&x + 1) I would be hard-pressed to call that idiomatic, however.
by caf 15y ago
You can also get the "correct" type (type of &x[0]) by using:
*(&x + 1)
I would be hard-pressed to call that idiomatic, however.