Once again: strict aliasing rule and char*

Question

The more I read, the more confused I get.

The last question from the related ones is closest to my question, but I got confused with all words about object lifetime and especially - is it OK to only read or not.

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To get straight to the point. Correct me if I'm wrong.

This is fine, gcc does not give warning and I'm trying to "read type T ( uint32_t ) via char* ":

uint32_t num = 0x01020304;
char* buff = reinterpret_cast< char* >( &num );

But this is "bad" (also gives a warning) and I'm trying "the other way around":

char buff[ 4 ] = { 0x1, 0x2, 0x3, 0x4 };
uint32_t num = *reinterpret_cast< uint32_t* >( buff );

How is the second one different from the first one, especially when we're talking about reordering instructions (for optimization)? Plus, adding const does not change the situation in any way.

Or this is just a straight rule, which clearly states: "this can be done in the one direction, but not in the other"? I couldn't find anything relevant in the standards (searched for this especially in C++11 standard).

Is this the same for C and C++ (as I read a comment, implying it's different for the 2 languages)?

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I used union to "workaround" this, which still appears to be NOT 100% OK, as it's not guaranteed by the standard (which states, that I can only rely on the value, which is last modified in the union ).

So, after reading a lot , I'm now more confused. I guess only memcpy is the "good" solution?

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Related questions:

• What is the strict aliasing rule?
• "dereferencing type-punned pointer will break strict-aliasing rules" warning
• Do I understand C/C++ strict-aliasing correctly?
• Strict aliasing rule and 'char *' pointers

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EDIT
The real world situation: I have a third party lib ( http://www.fastcrypto.org/ ), which calculates UMAC and the returned value is in char[ 4 ] . Then I need to convert this to uint32_t . And, btw, the lib uses things like ((UINT32 *)pc->nonce)[0] = ((UINT32 *)nonce)[0] a lot. Anyway.

Also, I'm asking about what is right and what is wrong and why. Not only about the reordering, optimization, etc. (what's interesting is that with -O0 there are no warnings, only with -O2 ).

And please note : I'm aware of the big/little endian situation. It's not the case here. I really want to ignore the endianness here. The "strict aliasing rules" sounds like something really serious, far more serious than wrong endianness. I mean - like accessing/modifying memory, which is not supposed to be touched; any kind of UB at all.

Quotes from the standards (C and C++) would be really appreciated. I couldn't find anything about aliasing rules or anything relevant.

Answer 1

How is the second one different from the first one, especially when we're talking about reordering instructions (for optimization)?

The problem is in the compiler using the rules to determine whether such an optimization is allowed. In the second case you're trying to read a char[] object via an incompatible pointer type, which is undefined behavior; hence, the compiler might re-order the read and write (or do anything else which you might not expect).

As unnatural as it might seem, you really have to stop thinking about how you think the compiler might optimize, and just obey the rules .

Or this is just a straight rule, which clearly states: "this can be done in the one direction, but not in the other"? I couldn't find anything relevant in the standards (searched for this especially in C++11 standard).

http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2012/n3337.pdf chapter 3.10 paragraph 10.

In C99, and I think also C11, it's 6.5 paragraph 7.

Both C and C++ allow accessing any object type via char * (or specifically, an lvalue of type char ). They do not allow accessing a char object via an arbitrary type. So yes, the rule is a "one way" rule.

I used union to "workaround" this, which still appears to be NOT 100% OK, as it's not guaranteed by the standard (which states, that I can only rely on the value, which is last modified in the union).

Although the wording of the standard is horribly ambiguous, in C99 (and beyond) it's clear (at least since C99 TC3) that the intent is to allow type-punning through a union. You must however perform all accesses through the union (in particular you cannot just 'cast a union into existence' for purpose of type-punning).

the returned value is in char[ 4 ]. Then I need to convert this to uint32_t

Just use memcpy or manually shift the bytes to the correct position, in case byte-ordering is an issue. Good compilers can optimize this out anyway (yes, even the call to memcpy ).

Answer 2

I used union to "workaround" this, which still appears to be NOT 100% OK, as it's not guaranteed by the standard (which states, that I can only rely on the value, which is last modified in the union).

Endianess is the reason for this. Specifically the sequence of bytes 01 00 00 00 could mean 1 or 16,777,216.

The correct way to do what you are doing is to stop trying to trick the compiler into doing the conversion for you and perform the conversion yourself.

For instance if the char[4] is little-endian (smallest byte first) then you would do something like the following.

char[] buff = new char[4];
uint32_t result = 0;
for (int i = 0; i < 4; i++)
    result = (result << 8) + buff[i];

This manually performs the conversion between the two and is guaranteed to always be correct as you are doing the mathematical conversion.

Now if you were doing this conversion rapidly it might make sense to use #if and knowledge of your architecture to use a enum to do this automatically as you mentioned, but that is again getting away from portable solutions. (Also you can use something like this as your fallback if you can't be certain)