How Java deal with hard coded numeric values?

Question

I have decompiled a Java class which contain this line :

DatabaseError.throwSqlException((int)23);

Until now i believe that Java consider that literal numeric value is int, So why they cast 23 ? And in general how Java deal with hard coded numeric values ?

Edit: Im not interested in this specific case, but interested to go deeper with Java in general.

Answer 1

I still believe it is related to the used decompiler.

It can be assumed that the signature of the method is DatabaseError.throwSqlException(int i).

The statement DatabaseError.throwSqlException((int)23) would be compiled as DatabaseError.throwSqlException(23) - the literal is of type int and the method parameter type is of type int

the DatabaseError.throwSqlException((int)23L) would be compiled as DatabaseError.throwSqlException(23) - the value can be stored without precision lose in an int and the method parameter type is of type int

The byte code is in both cases

bipush 23    // push the value on the operand stack
invokestatic // DatabaseError.throwSqlException:(I)V

Decompiled it would be DatabaseError.throwSqlException(23). Because in case it was 23L in the source code, this information isn't in the bytecode.

A bytecode similar to DatabaseError.throwSqlException((int)23) could be

ldc2_w       // push a long from constant pool on the operand stack
l2i          // convert top value on the operand stack from long to int 
invokestatic // DatabaseError.throwSqlException:(I)V

But this is decompiled by jad and jd-gui to DatabaseError.throwSqlException((int)23L) (notice the L after the value).

edit The above bytecode is decompiled by CFR as DatabaseError.throwSqlException((int)23). So it really depend on the used decompiler.