float bits and strict aliasing

问题

I am trying to extract the bits from a float without invoking undefined behavior. Here is my first attempt:

unsigned foo(float x)
{
    unsigned* u = (unsigned*)&x;
    return *u;
}

As I understand it, this is not guaranteed to work due to strict aliasing rules, right? Does it work if a take an intermediate step with a character pointer?

unsigned bar(float x)
{
    char* c = (char*)&x;
    unsigned* u = (unsigned*)c;
    return *u;
}

Or do I have to extract the individual bytes myself?

unsigned baz(float x)
{
    unsigned char* c = (unsigned char*)&x;
    return c[0] | c[1] << 8 | c[2] << 16 | c[3] << 24;
}

Of course this has the disadvantage of depending on endianness, but I could live with that.

The union hack is definitely undefined behavior, right?

unsigned uni(float x)
{
    union { float f; unsigned u; };
    f = x;
    return u;
}

Just for completeness, here is a reference version of foo. Also undefined behavior, right?

unsigned ref(float x)
{
    return (unsigned&)x;
}

So, is it possible to extract the bits from a float (assuming both are 32 bits wide, of course)?

EDIT: And here is the memcpy version as proposed by Goz. Since many compilers do not support static_assert yet, I have replaced static_assert with some template metaprogramming:

template <bool, typename T>
struct requirement;

template <typename T>
struct requirement<true, T>
{
    typedef T type;
};

unsigned bits(float x)
{
    requirement<sizeof(unsigned)==sizeof(float), unsigned>::type u;
    memcpy(&u, &x, sizeof u);
    return u;
}

回答1:

About the only way to truly avoid any issues is to memcpy.

unsigned int FloatToInt( float f )
{
   static_assert( sizeof( float ) == sizeof( unsigned int ), "Sizes must match" );
   unsigned int ret;
   memcpy( &ret, &f, sizeof( float ) );
   return ret;
}

Because you are memcpying a fixed amount the compiler will optimise it out.

That said the union method is VERY widely supported.

回答2:

The union hack is definitely undefined behavior, right?

Yes and no. According to the standard, it is definitely undefined behavior. But it is such a commonly used trick that GCC and MSVC and as far as I know, every other popular compiler, explicitly guarantees that it is safe and will work as expected.

回答3:

The following does not violate the aliasing rule, because it has no use of lvalues accessing different types anywhere

template<typename B, typename A>
B noalias_cast(A a) { 
  union N { 
    A a; 
    B b; 
    N(A a):a(a) { }
  };
  return N(a).b;
}

unsigned bar(float x) {
  return noalias_cast<unsigned>(x);
}

回答4:

If you really want to be agnostic about the size of the float type and just return the raw bits, do something like this:

void float_to_bytes(char *buffer, float f) {
    union {
        float x;
        char b[sizeof(float)];
    };

    x = f;
    memcpy(buffer, b, sizeof(float));
}

Then call it like so:

float a = 12345.6789;
char buffer[sizeof(float)];

float_to_bytes(buffer, a);

This technique will, of course, produce output specific to your machine's byte ordering.

来源：https://stackoverflow.com/questions/4328342/float-bits-and-strict-aliasing

标签

c++

floating-point

bits

strict-aliasing

type-punning