c++ - 缩小从无符号到 double 的转换

Question

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

g++ 4.7.1 提示此代码:

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

为什么这是一个缩小转换？不是每个 unsigned 都可以完美地表示为 double 吗？

Answer 1

Why is this a narrowing conversion?

因为定义包括(我强调):

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 不是常量表达式，因此无论源类型的所有可能值是否都可以在目标类型中表示，它都是一种缩小转换。

Isn't every unsigned perfectly representable as a double?

这是定义的实现。在常见的32位unsigned和double带52位尾数的情况下，就是这样；但有些实现有较大的 unsigned 和/或较小的 double 表示，因此依赖于该假设的代码是不可移植的。