c++ - narrowing conversion from unsigned to double

Question

static_assert(sizeof(unsigned) == 4, ":(");
static_assert(sizeof(double) == 8 ,":(");
unsigned u{42};
double x{u};

g++ 4.7.1 complains about this code:

warning: narrowing conversion of 'u' from 'unsigned int' to 'double' inside { }

Why is this a narrowing conversion? Isn't every unsigned perfectly representable as a double?

question from:https://stackoverflow.com/questions/11521016/narrowing-conversion-from-unsigned-to-double

Answer 1

Why is this a narrowing conversion?

Because the definition includes (with my emphasis):

C++11 8.5.4/7 A narrowing conversion is an implicit conversion [...] from an integer type [...] to a floating-point type, except where the source is a constant expression and the actual value after conversion will fit into the target type and will produce the original value when converted back to the original type.

u is not a constant expression, so it's a narrowing conversion whether or not all possible values of the source type might be representable in the target type.

Isn't every unsigned perfectly representable as a double?

That's implementation defined. In the common case of 32-bit unsigned and double with a 52-bit mantissa, that is the case; but some implementations have larger unsigned and/or smaller double representations, so code that depends on that assumption is not portable.