Performance The standard guarantees that if you do a default capture, the only variables that will be captured by that default capture from the surrounding environment are those that you actually use inside the lambda. As such, specifying individual variables to capture acts as documentation of what you expect to use, but should never affect … Read more
Java 8 introduces the Stream API that allows similar constructs to those in Linq. Your query for example, could be expressed: int cheetahNumber = 77; Animal cheetah = animals.stream() .filter((animal) -> animal.getNumber() == cheetahNumber) .findFirst() .orElse(Animal.DEFAULT); You’ll obviously need to workout if a default exists, which seems odd in this case, but I’ve shown it … Read more
What interests me is the Compile() method. Does it somehow produce real MSIL? Yes. The Compile method runs a visitor over the lambda body block and generates IL dynamically for each subexpression. If you’re interested in learning how to spit IL yourself, see this “Hello World” example of how to use Lightweight Codegen. (I note … Read more
Regular Ruby method calls use () not curly braces which are for blocks. If you don’t like [] for calling a lambda, you can always use the call method. Example: >> by_two = lambda { |x| x * 2 } #=> #<Proc:0x0000000101304588@(irb):1> >> by_two[5] #=> 10 >> by_two.call(5) #=> 10 Edit In newer version of … Read more
var query = context.ShoppingMalls .Join( context.Houses, s => new { s.CouncilCode, s.PostCode }, h => new { h.CouncilCode, h.PostCode }, (s, h) => s);
I think you are mixing two things here. Expression trees and delegates. Lambda can be used to express both of them, but it depends on the type of parameter the method accepts in which one it will be turned. A lambda passed to a method which as Action<T> or Func<T, TResult> will be converted into … Read more
Are there special rules that apply to constexpr for capturing/accessing? Yes, constexpr variables could be read without capturing in lambda: A lambda expression can read the value of a variable without capturing it if the variable has const non-volatile integral or enumeration type and has been initialized with a constant expression, or is constexpr and … Read more
The 2nd template parameter of std::set expects a type, not an expression, so it is just you are using it wrongly. You could create the set like this: auto comp = [](const A& lhs, const A& rhs) -> bool { return lhs.x < rhs.x; }; auto SetOfA = std::set <A, decltype(comp)> (comp);
Both seems to be desirable behavior for :if option mentioned above, is there anything I am missing? I’m guessing that: It’s more desirable to allow Procs as they don’t care about the number of arguments. So I could easily write any of the below: validates_presence_of :name, :if => Proc.new{|f| f.display_name.blank? } # I care about … Read more