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In what situations may `Number.toDouble()`  "involve rounding" as suggested by its API docs?

odd integers above 2^53 cannot be exactly represented as doubles

```&gt;&gt;&gt; (2L shl 53) + 1
res0: kotlin.Long = 18014398509481985
&gt;&gt;&gt; ((2L shl 53) + 1).toDouble().toLong()
res1: kotlin.Long = 18014398509481984```

<https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html>

Would be great to have this documented in Kotlin stdlib KDoc

does it have to be enumerated? javadoc uses the same phrasing

clearly the range of various numbers can't all fit in Double - there are other Number types such as BigInteger and BigDecimal as well

Thanks for the great answers. I agree, it does not seem to be "rounding"" if a `Number` value does not fit into the target type range, so some more details in the `Number` API docs would be great.

To be fair, most methods the `Number` docs state that it `may involve rounding or truncation.` which covers range issues. So the docs are somehow correct, although the big odd integers to double example is neither rounding nor truncation, so there is imho some  inprecision in the docs.

why wouldn't it be rounding? it rounds to a representable value

it just can't be called truncation, because when it runs out of finite values it goes infinite:
```2.toBigInteger().pow(1024).toDouble() == Double.POSITIVE_INFINITY```

That's a clver handing when exceeding the max value. And you're right, its rounding in technical sense that just takes some time to wrap ones head around it.