A DESCRIPTION OF THE PROBLEM :
Based upon the javadocs for BigDecimal, the lower bound for scale is -Integer.MAX_VALUE
Quote:
"The exponent consists of the character 'e' ('\u0065') or 'E' ('\u0045') followed by one or more decimal digits. The value of the exponent must lie between -Integer.MAX_VALUE (Integer.MIN_VALUE+1) and Integer.MAX_VALUE, inclusive."
Ref: https://docs.oracle.com/en/java/javase/13/docs/api/java.base/java/math/BigDecimal.html#%3Cinit%3E(java.lang.String)
If you do a .toString() on this value, I would expect to be able to deserialize the result using `new java.math.BigDecimal(str)`
java.math.BigDecimal.valueOf(Long.MIN_VALUE, -Integer.MAX_VALUE).toString => "-9.223372036854775808E+2147483665"
new java.math.BigDecimal("-9.223372036854775808E+2147483665") => Overflow Exception (because 2147483665 > Integer.MAX_VALUE of 2147483647) - and there is an (incorrect?) guard in place throwing the exception.
Via experimentation, the actual lower bound is `-Integer.MAX_VALUE+18` or `Integer.MIN_VALUE+19`:
java.math.BigDecimal.valueOf(Long.MIN_VALUE, -Integer.MAX_VALUE+18).toString = "-9.223372036854775808E+2147483647"
new java.math.BigDecimal("-9.223372036854775808E+2147483647").toString() => "-9.223372036854775808E+2147483647"
STEPS TO FOLLOW TO REPRODUCE THE PROBLEM :
import java.math.BigDecimal;
String str = java.math.BigDecimal.valueOf(Long.MIN_VALUE, -Integer.MAX_VALUE);
BigDecimal bd = new BigDecimal(str);
EXPECTED VERSUS ACTUAL BEHAVIOR :
EXPECTED -
It should not throw an exception.
FREQUENCY : always