compare void* with uint16_t

Question

In some legacy code, which compiles fine on GCC 4.6 (with -fpermissive), I have this:

uint16_t a = 0;
void* b = ...;

if(b == a) // ...

Is this comparison well-defined on GCC 4.6? Does it downcast to 16 bits or upcast to 32/64 bits?

If I remember correctly, this is valid in C but not in C++; I might be utterly wrong though... — Sebastian Hoffmann, Apr 04 '14 at 12:57
GCC will fail to compile this without the -fpermissive option. — NickC, Apr 04 '14 at 13:43

score 1 · Answer 1 · edited Jun 20 '20 at 09:12

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Although this isn't explicitly written in the C++11 standard (N3337 draft), I was able to come up with this (emphasis mine).

§5.9 Relational operators

Pointers to objects or functions of the same type (after pointer conversions) can be compared, with a result defined as follows

— If two pointers ...

— If two pointers ...

— If two pointers ...

— If two pointers ...

— If two pointers ...

— Other pointer comparisons are unspecified.

Now for the equality part:

§5.10 Equality operators

The == (equal to) and the != (not equal to) operators have the same semantic restrictions, conversions, and result type as the relational operators except for their lower precedence and truth-value result.

By this I believe that such a comparison is unspecified.

edited Jun 20 '20 at 09:12

Community

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answered Apr 04 '14 at 13:21

legends2k

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So you say that it is undefined in the C++ standard. But how would GCC 4.6 do it? – Albert Apr 04 '14 at 13:23
I believe that it is _unspecified_ by the standard, not _undefined_. However, in both cases an implemenation (like GCC) is free to do what it may; please see [here](http://stackoverflow.com/questions/2397984/undefined-unspecified-and-implementation-defined-behavior). – legends2k Apr 04 '14 at 13:23
I know. But my question is, what does it do. – Albert Apr 04 '14 at 13:43

score 1 · Accepted Answer · answered Apr 04 '14 at 13:50

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Looks like it up-casts the 16 bit integer to match the pointer size. Running the following code outputs "upcast"

uint16_t a = 1;
void* b = (void*)0x10001;
(b == a) ? printf("downcast") : printf("upcast");

answered Apr 04 '14 at 13:50

NickC

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Is that with GCC 4.6? – Albert Apr 04 '14 at 13:55
Thanks a lot! You should add that link to your answer. – Albert Apr 05 '14 at 11:51

score 0 · Answer 3 · answered Apr 04 '14 at 13:29

It may compile or may not (not sure, it depends on the compiler and compiler options) In any case the cast will be performed like:

if( b == (void*)a )
{
}

Note that upcasting/downcasting is not the correct word to use because it is related to classes, in this case is just type conversions.

compare void* with uint16_t

3 Answers3