缩小从无符号到双精度的转换 [英] narrowing conversion from unsigned to double

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

为什么这是缩小的转换？

Why is this a narrowing conversion?

因为定义包括b
$ b

Because the definition includes (with my emphasis):

C ++ 11 8.5.4 / 7 缩小转换次数是隐式转换
[...]从整数类型[...]更改为浮点类型，除非源是常量表达式，转换后的实际值将适合目标类型，

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.

不是每个无符号完美表示为 double ？

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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