array size dependent on sizeof() a struct field

Question

I have a global include file which contains a set of structures. Somewhere in my program, I have a class that contains a member array. The number of elements in this array is dependent on the size of a specific field in a specific struct. I want to make it so that the array size will get automatically updated if the sizeof the structure field is changed. I have been able to do this succesfully with the following expression:

bool shadowChkBox[sizeof(FSCconfigType::WriteEn)*8*MAX_FSCS];

FSCconfigType is the struct type and WriteEn is one of the fields. Now this worked but only on ubuntu. On RHEL 5, the compiler declared it as an error. What other alternatives could I have for doing this? I am working with Qt.

I suspect it has more to do with which version of GCC you are using, rather than which version of Linux. — xscott
– xscott, Commented Nov 28, 2010 at 21:50
Can you use the type of the field directly, rather than referencing it through the field? I'm guessing no (because that's a reason why the sizeof might change), but it doesn't hurt to check. — user180247
– user180247, Commented Nov 28, 2010 at 21:56
@steve in deed that is the reason why the sizeof might change — yan bellavance
– yan bellavance, Commented Nov 28, 2010 at 22:15
@yan bellavance, I agree that you shouldn't require your users to upgrade gcc. I was just trying to clarify the problem. I think the code should compile, and if it's a bug in the older version of gcc, then I think you're going to be stuck with an uglier workaround. — xscott
– xscott, Commented Nov 28, 2010 at 22:22

clstrfsck · Accepted Answer · 2010-11-28 23:27:11Z

2

Here is one possible answer:

#include <iostream>

struct A
{
        int a;
        int b;
        int c;
};

bool items[sizeof(reinterpret_cast<A *>(0)->b)];

int main()
{
        std::cout << sizeof(reinterpret_cast<A *>(0)->b) << ",";
        std::cout << sizeof(items) << std::endl;
        return 0;
}

answered Nov 28, 2010 at 23:27

clstrfsck

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2 Comments

yan bellavance Over a year ago

thx it worked. One question though. what exactly are you doing with the (0)? Seems to be an null address.

please delete me Over a year ago

The argument to sizeof is not evaluated except to determine its type. Casting NULL to A * is as good as a way as any to get something that has the type of pointer to A. (( A*)printf(0) would do just as well, for example.)

user180247 · Accepted Answer · 2010-11-28 21:59:32Z

0

One fragile answer is to take the difference of the offsetof values for WriteEn and the next field up (or failing that, the sizeof the whole struct).

This may give a slightly larger answer than sizeof due to alignment-based padding.

A bigger problem is what happens if you rearrange your fields without fixing this - but it might be an option if you're desperate.

answered Nov 28, 2010 at 21:59

user180247

Comments

Dmytro Ovdiienko · Accepted Answer · 2014-07-16 09:28:37Z

0

#include <iostream>

struct C
{
    int iv;
    void* pv;
    char buf[128];
};

template< typename TObj, typename TFieldType >
std::size_t field_size( TFieldType (TObj::*) )
{
    return sizeof(TFieldType);
}

int main() {

    std::cout << field_size(&C::iv) << std::endl;
    std::cout << field_size(&C::pv) << std::endl;
    std::cout << field_size(&C::buf) << std::endl;
}

answered Jul 16, 2014 at 9:28

Dmytro Ovdiienko

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