Reputation: 10132
I have a feeling the answer to this question is no, but is it possible to initialise a specific member of a union? For example the following:
#include <cassert>
#include <Windows.h>
int _tmain(int argc, _TCHAR* argv[])
{
auto time = 20090520145024798ull;
auto large = ULARGE_INTEGER() = {
{ time }
};
assert(large.QuadPart == time);
return 0;
}
(Visual Studio 2013, Windows 10), produces a compiler "conversion from 'unsigned __int64' to 'DWORD'", implying it's going to try to shoe-horn the uint64_t into the DWORD.
ULARGE_INTEGER is the union:
typedef union _ULARGE_INTEGER {
struct {
DWORD LowPart;
DWORD HighPart;
} u;
ULONGLONG QuadPart;
} ULARGE_INTEGER;
What does the standard say about order of initialisation in cases like this? I had hoped the compiler would see that QuadPart was the appropriate member to assign.
Upvotes: 1
Views: 121
Reputation: 3527
Keeping it simple:
ULARGE_INTEGER large;
large.QuadPart = time;
Or if you want to use auto and initialize it in one line, make a little helper perhaps?
inline auto MAKE_ULARGE_INTEGER(ULONGLONG t)
{
ULARGE_INTEGER result;
result.QuadPart = t;
return result;
}
auto large = MAKE_ULARGE_INTEGER(time);
Upvotes: 3
Reputation: 303517
With unions, you can use list-initialization to initialize the first member only, as per [dcl.init.aggr]:
When a union is initialized with a brace-enclosed initializer, the braces shall only contain an initializer-clause for the first non-static data member of the union. [ Example:
union u { int a; const char* b; }; u a = { 1 }; u b = a; u c = 1; // error u d = { 0, "asdf" }; // error u e = { "asdf" }; // error
—end example ]
So, in your example, the following:
ULARGE_INTEGER large{time};
would initialize u.LowPart
, not QuadPart
, regardless of what the types of the various members are.
If you want to do anything else, you'll have to be explicit about it:
ULARGE_INTEGER large;
large.QuadPart = time;
Or write a better union
type that actually has a constructor.
Upvotes: 4