2 ms·
When you are dealing with pointers (which is a requirement to implement the API of C's stdlib), you are in unsafe territory. Fortunately, any user of the C std
by tangent128 10y ago
When you are dealing with pointers (which is a requirement to implement the API of C's stdlib), you are in unsafe territory.
Fortunately, any user of the C stdlib (including other languages) already expects it to be unsafe.
- kbenson 10y ago> Fortunately, any user of the C stdlib (including other languages) already expects it to be unsafe. Yes, but it may be better to phrase it as "Fortunately, any user of the C stdlib already expects it to be unsafe by the standard and definition Rust uses". People often have their own idea of what unsafe is, and without making it explicitly clear, you might get people objecting to what is an obviously true statement purely because of a terminology mismatch.