Do data members form a range?

Question

Can I treat consecutive data members of the same type as a range? For example:

struct X
{
    int a, b, c, d, e;
};

X x = {42, 13, 97, 11, 31};

std::sort(&x.a, &x.a + 5);   // kosher?

Wasn’t there a guarantee somewhere about this for PODs (i.e. trivial and standard layout), where all the members have the same type, that they could be treated as arrays? — Konrad Rudolph, Sep 05 '13 at 13:22
Add a `static_assert(sizeof(X) == 5*sizeof(int), "I've been a really bad girl")`. — R. Martinho Fernandes, Sep 05 '13 at 15:19
`sizeof(X)` does not have to be exactly the size of 5 `int`s. If there is padding bytes at the end of the struct, the code could still do what it is expected to do. — Zyx 2000, Sep 05 '13 at 15:46
@Zyx2000 What would be the point of padding if all fields are of the same type? — fredoverflow, Sep 05 '13 at 16:46
@FredOverflow I don't know any particular reason, but I believe implementations are allowed to do that. — Zyx 2000, Sep 05 '13 at 21:54
@R.MartinhoFernandes I meant that `sizeof(X)` does not have to be 5 `int`s for this trick to work. — Zyx 2000, Sep 05 '13 at 21:55

score 15 · Accepted Answer · answered Sep 05 '13 at 12:17

15

No, this is undefined behaviour. You are treating x.a like the first element of an array, which it isn't. May work on some implementations, may raid your fridge too ;)

answered Sep 05 '13 at 12:17

Paul Evans

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Correct, but `For the purposes of these operators, a pointer to a nonarray object behaves the same as a pointer to the first element of an array of length one with the type of the object as its element type.` So you can treat it as a pointer to the first element of an array, just the length is one. – Jesse Good Sep 05 '13 at 12:20
2

@JesseGood That quote allows `X* p = &x; ++p;` to get a pointer past the end, but not what's in OP. – jrok Sep 05 '13 at 12:27
@jrok But for completeness: once you get said pointer to one past the end you can use it to refer to the second element *if it is there*. C++ doesn't care whence you get the pointer. Of course, if it's not there, you get the nasties. – R. Martinho Fernandes Sep 05 '13 at 15:13
@R.MartinhoFernandes I'm not sure what you mean. Is it there? Or is it not there? – fredoverflow Sep 05 '13 at 16:47

score 7 · Answer 2 · answered Sep 05 '13 at 12:23

7

Don't do that. Compiler is free to add paddings between structure members(and at the end).

answered Sep 05 '13 at 12:23

Eric Z

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What would be the point of padding if all fields are of the same type? – fredoverflow Sep 05 '13 at 16:48
@FredOverflow faster access in case the CPU does not have specialized load/store instructions or alignment restrictions. – josefx Sep 15 '13 at 11:05

score 4 · Answer 3 · answered Sep 05 '13 at 12:52

If this is really something you want to do, make it an array, vector or similar.

As others have said, the standard makes no guarantees about the members being stored without gaps or otherwise things that cause problems. (And to make matters worse, it will appear to work, until you compile it with a different (version of) compiler, or for another architecture some months or years later, and of course, it won't be easy to figure out what went wrong).

score 2 · Answer 4 · answered Sep 05 '13 at 12:17

2

No, this is not possible in C++. It would be rather difficult to standarize; in this case the semantic is simple enough, but what if the struct was heterogenous?

answered Sep 05 '13 at 12:17

Marcin Łoś

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score 0 · Answer 5 · answered Sep 07 '13 at 21:21

0

Generally, it's a bad idea. If you want to treat some variables as array, you should declare them as array :)

Nevertheless you can use some compiler-specific instruction to ensure that there is no padding between elements, f.i.:

#pragma pack(push, 1)  
struct X
{
    int a, b, c, d, e;
};
#pragma pack(pop)

or

struct  __attribute__((__packed__)) X
{
    int a, b, c, d, e;
};

answered Sep 07 '13 at 21:21

nevermind

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Why would there be padding between integers (or any other data member of one type)? – fredoverflow Sep 08 '13 at 11:20

Do data members form a range?

5 Answers5