blocks|key|2469693|text|可以由float或double表示的值的范围要比由long表示的范围大得多。虽然在从long转换到float时可能会丢失重要的数字，但由于范围更宽，它仍然是一个“扩大”操作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2469694|来自Java语言规范，第5.1.2节|2469695|若将int或长值转换为浮动，或将长值转换为双倍，则可能会导致精度下降--也就是说，结果可能会丢失一些最不重要的值。在这种情况下，产生的浮点值将是整数值的正确舍入版本，使用IEEE+754轮到最近模式(§4.2.4)。|blockquote|2469696|注意，double可以准确地表示每个可能的int值。|2469697|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.1.2^0|3|5|9|6|P|4|16|4|1D|5|0|2|G|0|0|0|3|6|L|3|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|15|8|@]|D|@$9|16|A|17|1|18]]|E|$]]|$1|H|3|I|5|J|7|19|8|@]|D|@]|E|$]]|$1|K|3|L|5|6|7|1A|8|@$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

The range of values that can be represented by a <code>float</code> or <code>double</code> is much larger than the range that can be represented by a <code>long</code>. Although one might lose significant digits when converting from a <code>long</code> to a <code>float</code>, it is still a "widening" operation because the range is wider.

From the <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.1.2" rel="noreferrer">Java Language Specification, §5.1.2</a>:

<blockquote>
 A widening 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).
</blockquote>

Note that a <code>double</code> can exactly represent every possible <code>int</code> value.

blocks|key|2469703|text|因为float和double可以表示比long更大的值，所以它被认为是扩展的。您可能会丢失精度，但表示值(至少大约)是可能的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2469704|entityMap^0|2|5|8|6|J|4|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

It's considered widening because <code>float</code> and <code>double</code> can represent larger values than <code>long</code>. You may lose precision, but it will be possible to represent the value (at least approximately).

blocks|key|1680126|text|它被认为是扩大化的，因为可以用浮点数表示的数字比用long表示的数字要大。仅仅因为浮点数使用32位精度并不意味着它可以表示的数字限制在2%5E32。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1680127|例如，浮动(float)Long.MAX_VALUE%2B(float)Long.MAX_VALUE大于Long.MAX_VALUE，尽管浮点数的精度低于long。|offset|length|style|CODE|1680128|entityMap^0|0|5|17|1E|E|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]|$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|I|$]]

It is considered widening because the numbers that can be represented by a float is larger than numbers that can represented by long. Just because float uses 32 bit precision does not mean the numbers it can represent are limited to 2^32.

For instance the float <code>(float)Long.MAX_VALUE+(float)Long.MAX_VALUE</code> is larger than <code>Long.MAX_VALUE</code>, even though the float has less precision that the long.

blocks|key|2469773|text|如果您简单地看待这个问题，那就是数据是如何由原始设计人员表示的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2469774|理想情况下，长(64)位深度大于浮子(32)。但是浮点数据是用科学观念表示的，这使得表示范围要大得多。|2469775|例:300个原始数字:+3×102科学代表|2469776|长：-2%5E63至2%5E63-1|2469777|浮动：(-3.4)*10%5E38至(3.4)*10%5E38|2469778|注意，在这里，长(2)v的幂(10的幂)代表差异，这允许浮动有更高的范围。|2469779|希望这是有帮助的|2469780|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Scientific_notation^0|G|E|0|V|4|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|12|8|@]|D|@$9|13|A|14|1|15]]|E|$]]|$1|H|3|I|5|6|7|16|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|17|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|18|8|@]|D|@]|E|$]]|$1|N|3|O|5|6|7|19|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1A|8|@]|D|@]|E|$]]|$1|R|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|S|$T|$5|U|V|W|E|$X|Y]]]]

If you look at the matter in simple terms, it is about how data has been represented by original designers.

ideally bit depth of long(64) is larger than float(32). But float data has represented using <a href="https://en.wikipedia.org/wiki/Scientific_notation" rel="nofollow noreferrer">scientific notion</a>
which allows to represent considerably much larger range

Ex: 300[original number] : 3×102 [scientific representation] 

<hr>

Long : -2^63 to 2^63-1

Float : (-3.4)*10^38 to (3.4)*10^38

Notice the Long(power of two) Vs Float(power of ten) representational difference here which allow float to have higher range 

hope this is helpful

As it states from oracle

Reference from <a href="http://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html" rel="noreferrer">Oracle Docs</a>

Widening Primitive Conversion
19 specific conversions on primitive types are called the widening primitive conversions:

<ol>
<li>byte to short, int, long, float, or double</li>
<li>short to int, long, float, or double</li>
<li>char to int, long, float, or double</li>
<li>int to long, float, or double</li>
<li>long to float or double?</li>
<li>float to double</li>
</ol>

If a float has 32 bits and a long has 64 how is that considered widening? Shouldn't this be considered narrowing?

Java Why is converting a long (64) to float (32) considered widening?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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，正如它从oracle中声明的那样参考资料对基元类型的19种特定转换称为加宽原语转换：字节到短、int、长、浮点或双字节短到整，长，浮动或双倍长的，浮动的，双倍的整数到长、浮动或双倍长到浮动或双浮到双倍如果浮点数有32位，而长有64位，这怎么被认为是加宽呢？这难道不应该被认为是缩小范围吗？

问Java为什么考虑将一个长(64)到浮点数(32)转换为加宽？
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