如何将位域成员的所有位设置为1

Question

The question is pretty simple. Given

struct Foo{
    bool     : 1;
    bool     : 1;
    int bar  : sizeof(int) * 8 - 2;
};

How can I set all bits of bar to 1, without a warning?

Obviously I can do auto v = Foo(); v.bar = ~0; auto v = Foo(); v.bar = ~0; but GCC gives me this:

warning: large integer implicitly truncated to unsigned type [-Woverflow]

I have tried a couple of ways, but it always renders a warning...

Answer 1

Don't use ~0 (which is always a full int ), but ~v.bar , which is the right size and can be combined like so:

v.bar |= ~v.bar; // or,
v.bar ^= ~v.bar;

should do the trick, for any size.

Unfortunately you can't nicely wrap it in a function since you can't bind references to bitfields. You'd either need to make it a function on Foo& , or use a macro.

PS. I tried this quickly before posting with coliru , and just re-checked locally with GCC (g++) 5.3.1 and 6.2.0 - neither issue any diagnostic with -Wall .

PPS.

With this test code, GCC produces a diagnostic only for the unsigned member:

struct Foo {
  int i : 2;
  unsigned int u : 30;
};

void bar() {
  Foo f {0, 0};
  f.i ^= ~f.i; // OK
  f.u |= ~f.u; /* warning:
    large integer implicitly truncated to unsigned type [-Woverflow]
  */
}

So although I'd also usually prefer unsigned integers for bitfields or for bitwise manipulation, g++ is happier and quieter with int here.

Answer 2

This works for me for unsigned types. Note that all bit-fiddling should be done with unsigned types, unless you're OK with undefined behaviour.

...
   unsigned int bar  : sizeof(int) * 8 - 2;
...

template <typename T, unsigned n>
constexpr T onebits()
{
    return 1 | (((1 << (n-2)) - 1) << 1);
}

v.bar = onebits<unsigned, sizeof(int) * 8 - 2>();