Timeline for answer to Reinterpret_cast vs. C-style cast by Davislor

7 events

when toggle format	what		by	license	comment
Jul 17, 2018 at 13:23	history	edited	Davislor	CC BY-SA 4.0	added 32 characters in body
Jul 17, 2018 at 13:10	comment	added	curiousguy		Yes indeed a `reinterpret_cast` to a reference type does a reinterpretation. That a reference type was used wasn't clear from your previous comment.
Jul 16, 2018 at 18:40	comment	added	Davislor		@curiousguy [expr.reinterpret.cast.11], although that says it needs to be cast to a reference.
Jul 16, 2018 at 17:32	comment	added	Davislor		@curiousguy Certain integer casts do. A cast between signed and unsigned integral types does, and so does a cast from `void*` to `uintptr_t`. On the other hand, a `reinterpret_cast` from `double` to `uint64_t` is a type-pun, while a C-style cast has the semantics of `static_cast`, which represents the value as closely as possible.
Jul 16, 2018 at 16:23	comment	added	Davislor		@curiousguy That’s when you reinterpret the bits of some object’s binary representation as if it were a different type. For example, treating the bits of a 64-bit `double` as a `uint64_t` so you can twiddle them. Especially when the method is declaring a `union`, writing to a member of one type, and reading out a member of a different type (which is legal in C, but, as you know, undefined behavior in C++).
Jul 16, 2018 at 15:07	comment	added	curiousguy		What is a "type-pun"?
Jul 9, 2018 at 18:57	history	answered	Davislor	CC BY-SA 4.0