Communities

Writing
Writing
Codidact Meta
Codidact Meta
The Great Outdoors
The Great Outdoors
Photography & Video
Photography & Video
Scientific Speculation
Scientific Speculation
Cooking
Cooking
Electrical Engineering
Electrical Engineering
Judaism
Judaism
Languages & Linguistics
Languages & Linguistics
Software Development
Software Development
Mathematics
Mathematics
Christianity
Christianity
Code Golf
Code Golf
Music
Music
Physics
Physics
Linux Systems
Linux Systems
Power Users
Power Users
Tabletop RPGs
Tabletop RPGs
Community Proposals
Community Proposals
tag:snake search within a tag
answers:0 unanswered questions
user:xxxx search by author id
score:0.5 posts with 0.5+ score
"snake oil" exact phrase
votes:4 posts with 4+ votes
created:<1w created < 1 week ago
post_type:xxxx type of post
Search help
Notifications
Mark all as read See all your notifications »
Q&A

Welcome to Software Development on Codidact!

Will you help us build our independent community of developers helping developers? We're small and trying to grow. We welcome questions about all aspects of software development, from design to code to QA and more. Got questions? Got answers? Got code you'd like someone to review? Please join us.

Is `uint8_t` always an alias for a character type if it exists?

+8
−0

Is uint8_t guaranteed to be a character type if it exists? Will using a uint8_t* to examine bytes of an object cause violation of the strict aliasing rule? Is the following legal code:

#include <cstddef>
#include <cstdint>
#include <cstdio>
#include <string>

int main() {
  std::string str{"Hello"};
  std::uint8_t* p = reinterpret_cast<std::uint8_t*>(&str);
  for (std::size_t i = 0; i < sizeof str; ++i) {
    std::printf("%d\n", *p++);
  }
}
History
Why does this post require moderator attention?
You might want to add some details to your flag.
Why should this post be closed?

0 comment threads

1 answer

+6
−0

Yes, it is in practice always a character type and you can safely assume as much, both in terms of (g)lvalue access and in terms of strict pointer aliasing. If not, the compiler would soon render itself completely useless.

C and C++ both got the following rule (C17 7.20.1.1/3)

intN_t ... uintN_t ...

These types are optional. However, if an implementation provides integer types with widths of 8, 16, 32, or 64 bits, no padding bits, and (for the signed types) that have a two’s complement representation, it shall define the corresponding typedef names.

So if your system supports 8 bit 2's complement numbers, it must support uint8_t. No exceptions - not even for freestanding (embedded) systems - stdint.h is one of the mandatory headers for all conforming implementations (C17 4/6).

And for such a system it does not make sense to define unsigned char as anything else but 8 bits. CHAR_BITS will be 8.

Padding bits, trap representations and other such exotic oddities is not allowed for character types either, nor can trap representations exist in 2's complement integers.

In practice, all known real-world compilers will simply implement uint8_t as a typedef for unsigned char. You can easily prove this by trial and error:


C

_Generic((uint8_t){0}, uint8_t:0, unsigned char:0);

error: '_Generic' specifies two compatible types


C++

void f (unsigned char c){}
void f (uint8_t c){}

error: redefinition of void f(uint8_t)
note: void f(unsigned char) previously defined here void f (unsigned char c){}


For those few exotic systems that have 16 bit bytes or other oddities like 1's complement, they cannot support uint8_t in the first place.

History
Why does this post require moderator attention?
You might want to add some details to your flag.

1 comment thread

General comments (6 comments)

Sign up to answer this question »