gortiz commented on code in PR #14337:
URL: https://github.com/apache/pinot/pull/14337#discussion_r1832331941
##########
pinot-core/src/main/java/org/apache/pinot/core/operator/transform/function/JsonExtractScalarTransformFunction.java:
##########
@@ -184,8 +191,7 @@ public long[] transformToLongValuesSV(ValueBlock
valueBlock) {
if (result instanceof Number) {
_longValuesSV[i] = ((Number) result).longValue();
} else {
- // Handle scientific notation
- _longValuesSV[i] = (long) Double.parseDouble(result.toString());
+ _longValuesSV[i] = Long.parseLong(result.toString());
Review Comment:
> JSON standard does allow that for numbers but not for integers.
That is not precise. In JSON there is only one type of numeric literal:
_number_. A _number_ is defined as `number fraction exponent`. Therefore you
can write a number as `9007199254740993e0` and it will generate the _number_
`9007199254740993e`. We have to interpret that as `9007199254740992`, but
`(long) Double.parseDouble("9007199254740993e0")` returns `9007199254740992`.
Using `BigInteger`/`BigDecimal` should be correct, but IMHO pretty
expensive. I think it shouldn't be that hard to write our own parser. Something
like:
```java
int e = indexOfExponent(str);
if (e < 0) {
return Long.parse(str);
}
long base = Long.parse(str, 0, e, 10));
int exp = Integer.parseInt(str, e+1, 10));
long powerOfTwo = powerOfTwo(exp); // as suggested in
https://stackoverflow.com/questions/46983772/fastest-way-to-obtain-a-power-of-10
long result = verifyOverflow(base * powerOfTen(exp)); // Not sure if
checking if the value changed the sign is good enough
```
Alternatively, if we are not sure if that algorithm is correct, we can do:
```java
int e = indexOfExponent(str);
if (e < 0) {
return Long.parse(str);
}
if (str.contains('.')) {
throw whatever;
}
return toLong(new BigDecimal(str)); // verifying we fail if the value is not
representable as double.
```
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