How to get float bit representation without UB in C++?

Question

As far as I know, all 'traditional' ways of doing this, namely reinterpret_cast of a pointer and union with int and float fields are UB as violation of strict aliasing (in C++, not in C).
So, how to do it correctly without undefined behavior?

Can I do a reinterpret_cast to char * and memcpy that to uint32_t? Or maybestd::launder will help?

If all you want to do is “get” the bit representation, you can simply convert a pointer to the `float` to `const unsigned char *` and use the resulting pointer to examine the bytes of the object. But why do you mention “memcpy that to float”? Copying to a `float` is not getting the bit representation; it is changing the bits. What do you really want to do? — Eric Postpischil, Oct 13 '19 at 20:06
There was an [1h talk](https://www.youtube.com/watch?v=_qzMpk-22cc) about this problem on cppcon 2019. You might be interested in it. — Timo, Oct 13 '19 at 20:06
@EricPostpischil yeah, sorry, I meant 'memcpy to uint32_t`, I'll edit the question. — Amomum, Oct 13 '19 at 20:07
`memcpy that to uint32_t` - why would you do that to print bits? You can print them from `char`. — KamilCuk, Oct 13 '19 at 20:08
@KamilCuk well, I want to have them as a single continuous thingy, to get significand and exponent; getting them from char array will be a bit tedious. — Amomum, Oct 13 '19 at 20:11

Resurrection · Accepted Answer · 2019-10-13T20:19:00.463

2

The standard way to do it, as pointed out by Jason Turner, is to use memcpy:

float f = 1.0;
std::byte c[sizeof(f)];
memcpy(c, &f, sizeof(f));

You may be thinking that you do not want to copy anything, that you just want to see the bits/bytes. Well the compilers are smart and they will in fact optimize it away as demonstrated by Jason so do not worry and use memcpy for this kind of thing and never reinterpret_cast.

edited Oct 13 '19 at 20:19

answered Oct 13 '19 at 20:08

Resurrection

3,916
2
34
56

4

C++20 [`std::bit_cast()`](https://en.cppreference.com/w/cpp/numeric/bit_cast) deserves a mention here. – Deduplicator Oct 13 '19 at 20:10
Exactly! That is going to be THE standard way of doing this. – Resurrection Oct 13 '19 at 20:11
I'm not afraid of copies, I'm afraid of undefined behavior :) Thanks! And what about `std::launder`? – Amomum Oct 13 '19 at 20:12
@Amomum `std::laundr`s return type is the same as its argument type. So conversion from `float` to `unsigned` is not possible. – Timo Oct 13 '19 at 20:17
@Timo what about `std::launder(reinterpret_cast(f))`? – Amomum Oct 13 '19 at 20:19
@Amomum why use `laundr` in the first place then? To quote [cppreference](https://en.cppreference.com/w/cpp/utility/launder) on `std::laundr`: *"the type of X is the same as T"* (where X is the object pointed to). – Timo Oct 13 '19 at 20:23
@Timo I'm just trying to understand if using `launder` will make such cast or union usage legal. – Amomum Oct 13 '19 at 20:25
@Amomum - It won't. That's not the intended use case, and digging into the wording leaves it still undefined to use the pointer. – StoryTeller - Unslander Monica Oct 13 '19 at 21:28
@Amomum - Still no. `launder` deals only with objects within their lifetime. In a union, the object you *want* is not within its lifetime. – StoryTeller - Unslander Monica Oct 14 '19 at 08:56

How to get float bit representation without UB in C++?

1 Answers1