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Casting between different sized integer references

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If you cast references between two integers of different sizes, say using reinterpret_cast, you start sinking into dangerous undefined behavior, right?

uint16_t originalInt = 5000;

uint8_t &smallerThanOriginalRef = reinterpret_cast<uint8_t&>(originalInt);
uint32_t &largerThanOriginalRef = reinterpret_cast<uint32_t&>(originalInt);

Would assigning to 'smallerThanOriginalRef' only set the first byte of 'originalInt'?

So this:

smallerThanOriginalRef = 123;

Wouldn't first clear the higher bytes of 'originalInt', and so 'originalInt' wouldn't be guaranteed to be 123, right?

 

And assigning to 'largerThanOriginalRef' would accidentally write on bytes outside of 'originalInt', which might write on memory that is used by other variables?

 

(in practice, those ints might be internally represented as a 32 bit or 64 bit integer anyway... but that's not guaranteed by the standard)

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Yup, undefined behavior territory. In practice what would happen depends on what endianness the platform is. Assigning to smallerThanOriginalRef might end up clearing either the higher or lower byte of originalInt.

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If you are sizing it up use the & to clean out the upper bits. Also you don't have to use reinterpret_cast, takes longer to type out than simply using v=(uint32_t)variable, which will do the exact same thing.

 

uint32_t largerThanOriginalRef = ((uint32_t)originalInt)&0xFFFF;

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Also you don't have to use reinterpret_cast, takes longer to type out than simply using v=(uint32_t)variable, which will do the exact same thing.


And also has the wonderful benefit of making your evil cast impossible to find in a simple text search!

We should always favor saving a couple of keystrokes to make our lives more miserable down the road.

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Also you don't have to use reinterpret_cast, takes longer to type out than simply using v=(uint32_t)variable, which will do the exact same thing.

 

My original use had to do with passing a uint32_t to a function wanting a uint8_t&, but then I realized the all problems involved with that. laugh.png

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You would lose a huge amount of precision converting it to an eight bit value. If you really needed it 8 bit then extract it out in the routine itself, c++ normally passes everything as 32 bits anyways. Anything larger (such as strings) I always pass as a pointer reference using &... The only case I avoid doing this is for logging where I might send in a full string such as Log("Feature xxxx failed to work"); otherwise it is always a variable.

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Interestingly, GCC 4.8 does not even warn about that code of yours although it's arguably in violation of the standard which says "A reference shall be initialized to refer to a valid object or function" (8.3.2) with "valid" being the important bit.

 

Since originalInt is not of a type that the new reference type can accomodate, it isn't a valid object (well, originalInt itself is a valid object, but result of the cast which the reference is initialized with isn't). You would think that this is obvious to the compiler, too. But maybe it's because of the cast operation. Probably the compiler assumes "programmer said cast, so he knows what he's doing".

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It would be the same as if pointers were used.

uint16_t originalInt = 5000;
uint8_t &smallerThanOriginalRef = reinterpret_cast<uint8_t&>(originalInt);
uint32_t &largerThanOriginalRef = reinterpret_cast<uint32_t&>(originalInt);
 
smallerThanOriginalRef = 123;
largerThanOriginalRef = 1000000;
Should behave the same as:

uint16_t originalInt = 5000;
uint8_t *smallerThanOriginalPtr = reinterpret_cast<uint8_t*>(&originalInt);
uint32_t *largerThanOriginalPtr = reinterpret_cast<uint32_t*>(&originalInt);
 
*smallerThanOriginalPtr = 123;
*largerThanOriginalPtr = 1000000;
The smaller write would right to the beginning bits of the original int, and whether that's the low or high bits depends on endienness. The larger would write beyond the memory of the original int, possibly clobbering another stack variable.

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Unspecified behavior not undefined behavior. 

 

Undefined behavior is always wrong code.

Unspecified behavior means it is defined by the implementation.

Edited by Shannon Barber

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Interestingly, GCC 4.8 does not even warn about that code of yours although it's arguably in violation of the standard which says "A reference shall be initialized to refer to a valid object or function" (8.3.2) with "valid" being the important bit.

 

Since originalInt is not of a type that the new reference type can accomodate, it isn't a valid object (well, originalInt itself is a valid object, but result of the cast which the reference is initialized with isn't). You would think that this is obvious to the compiler, too. But maybe it's because of the cast operation. Probably the compiler assumes "programmer said cast, so he knows what he's doing".

He cast a POD to a POD so it's valid as long memory (size) constraints are honored and they are since it's a smaller.

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Unspecified behavior not undefined behavior.

You're probably looking at the section of the standard regarding reinterpret_cast which states that the pointer value of the reinterpret_cast is unspecified. However, you're still modifying an object through an lvalue that is different than its dynamic type (and doesn't differ only in cv-qualifiers or signed/unsigned, isn't part of an aggregate, etc.), which is undefined behavior. See section 3.10 in the standard (all versions). (With the exception that uint8_t is probably a typedef for unsigned char, and modifying an object through an lvalue of char or unsigned char isn't undefined behavior. The largerThanOriginal is definitely undefined since there's no way for it to be a typedef for char or unsigned char and still have the example compile.)

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