What you have here, is a follow-up error which masks the original error. When lambda instances are serialized, they use writeReplace to dissolve their JRE specific implementation from the persistent form which is a SerializedLambda instance. When the SerializedLambda instance has been restored, its readResolve method will be invoked to reconstitute the appropriate lambda instance. … Read more
What you have here, is a follow-up error which masks the original error. When lambda instances are serialized, they use writeReplace to dissolve their JRE specific implementation from the persistent form which is a SerializedLambda instance. When the SerializedLambda instance has been restored, its readResolve method will be invoked to reconstitute the appropriate lambda instance. … Read more
A let is a lambda. E.g. (let ((x 1)) body) can be translated into ((lambda (x) body) 1) Furthermore, in Scheme all control and environment structures can be represented by lambda expressions and applications of lambdas. So, lambda is strictly more powerful than let and forms the basis of many of the interesting constructs found … Read more
The issue has nothing to do with thread safety, really. There’s a simple, straightforward answer to why instance variables can always be captured: this is always effectively final. That is, there is always one known fixed object at the time of the creation of a lambda accessing an instance variable. Remember that an instance variable … Read more
This has just been fixed in gcc-4.7 trunk and gcc-4.6 branch. These should be available in gcc-4.7.0 (a while from now – still in stage 1) and gcc-4.6.2 (alas 4.6.1 just came out.) But the intrepid could wait for the next snapshots or get a subversion copy. See audit trail for details.
This line is definitely ambiguous: doStuff(getPattern(x -> String.valueOf(x))); Reread this from the linked JLS chapter: A lambda expression (§15.27) is potentially compatible with a functional interface type (§9.8) if all of the following are true: The arity of the target type’s function type is the same as the arity of the lambda expression. If the … Read more
Nope, these are the two ways of doing it. Anything else would end up being only less clear. But, since you asked, here are some options. static<T,U,R> Function<T,R> chain( Function<? super T, ? extends U> f1, Function<? super U, ? extends R> f2) { return t -> f2.apply(f1.apply(t)); } stream.map(chain(Status::getUser, User::getId)) Or static<T,R> Function<T,R> func(Function<T,R> … Read more
Thanks for reminding me of how pointless this feature is. The correct answer is: no, that’s not a well-formed lambda. The grammar is defined in [expr.prim.lambda.general]: In our case, to start with we have: []<auto> noexcept requires true {}; [] is the lambda-introducer <auto> matches <template-parameter-list> and now we know we’re the 2nd kind lambda-expression. … Read more
The code to create the expression dynamically would be like this: ParameterExpression parameter = Expression.Parameter(typeof (GoalsModelUnitOfWork), “i”); MemberExpression property = Expression.Property(parameter, “AcademicCycles”); var queryableType = typeof (IQueryable<>).MakeGenericType(typeof (AcademicCycle)); var delegateType = typeof (Func<,>).MakeGenericType(typeof (GoalsModelUnitOfWork), queryableType); var yourExpression = Expression.Lambda(delegateType, property, parameter); The result will have the desired type, but the problem is that the return … Read more