问题描述
static_assert(sizeof(unsigned) == 4, ":(");
static_assert(sizeof(double) == 8 ,":(");
unsigned u{42};
double x{u};
g ++ 4.7。 1抱怨这个代码：
g++ 4.7.1 complains about this code:
warning: narrowing conversion of 'u' from 'unsigned int' to 'double' inside { }
为什么这是一个缩小的转换？不是每个无符号完美表示为 double ？
Why is this a narrowing conversion? Isn't every unsigned perfectly representable as a double?
推荐答案
因为定义包括b
$ b
Because the definition includes (with my emphasis):
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，它是一个缩小的转换。
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.
这是实现定义。在32位 unsigned 和 double 与52位尾数的通用情况下，情况是这样;但是一些实现具有更大的 unsigned 和/或更小的 double 表示，因此依赖于该假设的代码是不可移植的。
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.
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