关于转换：为什么Java隐式（没有强制转换）将`long`转换为`float`？

Why does Java implicitly (without cast) convert a `long` to a `float`?

每次我认为我了解铸造和转化，我发现另一个奇怪的行为。

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long l = 123456789L;
float f = l;
System.out.println(f); // outputs 1.23456792E8

考虑到long比float具有更大的位深度，我希望需要显式的强制转换才能编译它。不足为奇，我们看到我们在结果中失去了精确性。

为什么这里不需要演员表？

同样的问题也可以问到long到double-两种转换都可能丢失信息。

Java语言规范的第5.1.2节说明：

Widening primitive conversions do not
lose information about the overall
magnitude of a numeric value. Indeed,
conversions widening from an integral
type to another integral type do not
lose any information at all; the
numeric value is preserved exactly.
Conversions widening from float to
double in strictfp expressions also
preserve the numeric value exactly;
however, such conversions that are not
strictfp may lose information about
the overall magnitude of the converted
value.

Conversion of an int or a long value
to float, or of a long value to
double, may result in loss of
precision-that is, the result may lose
some of the least significant bits of
the value. In this case, the resulting
floating-point value will be a
correctly rounded version of the
integer value, using IEEE 754
round-to-nearest mode (§4.2.4).

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换句话说，即使您可能会丢失信息，您也知道该值仍将在目标类型的整个范围内。

当然，我们可以选择要求所有的隐式转换完全不丢失任何信息，因此，int和long到float应该是显式的，long到double应该是显式的。(int到double是可以的；double有足够的精度精确地表示所有int值。)

在某些情况下是有用的——在某些情况下不是。语言设计是关于妥协的，你不能全部取胜。我不知道我会做出什么决定…

相关讨论

Java语言规范，第5章：转换和推广解决了这个问题：

5.1.2 Widening Primitive Conversion

The following 19 specific conversions
on primitive types are called the
widening primitive conversions:

byte to short, int, long, float, or double

short to int, long, float, or double

char to int, long, float, or double

int to long, float, or double

long to float or double

float to double

Widening primitive conversions do not lose information about the overall magnitude of a numeric value.

...

Conversion of an int or a long value to float, or of a long value to double, may result in loss of precision-that is, the result may lose some of the least significant bits of the value. In this case, the resulting floating-point value will be a correctly rounded version of the integer value

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换句话说，JLS区分了量级损失和精度损失。

例如，int到byte是一个(潜在的)数量级损失，因为你不能在byte中存储500个。

long到float是一个潜在的精度损失，但不是数量级，因为浮动的值范围大于长的值范围。

所以规则是：

损失量：需要明确的铸造；
精度损失：无需铸造。

微妙？当然。但我希望能澄清这一点。