In c++14 using static_cast / const_cast: [&best_string = static_cast<const std::string&>(best_string)](const string& s) { best_string = s; // fails }; DEMO In c++17 using std::as_const: [&best_string = std::as_const(best_string)](const string& s) { best_string = s; // fails }; DEMO 2
int maxShoeSize = Workers.Where(x => x.CompanyId == 8) .Select(x => x.ShoeSize) .DefaultIfEmpty(0) .Max(); The zero in DefaultIfEmpty is not necessary.
UPDATE: This answer is in response to the original question, Does Java SE 8 have Pairs or Tuples? (And implicitly, if not, why not?) The OP has updated the question with a more complete example, but it seems like it can be solved without using any kind of Pair structure. [Note from OP: here is … Read more
This was explained by Mads Torgersen in C# Design Meeting Notes where local functions were first discussed: You want a helper function. You are only using it from within a single function, and it likely uses variables and type parameters that are in scope in that containing function. On the other hand, unlike a lambda … Read more
In IntelliJ IDEA: In File Menu → Project Structure → Project, change Project Language Level to 8.0 – Lambdas, type annotations etc. For Android 3.0+ Go File → Project Structure → Module → app and In Properties Tab set Source Compatibility and Target Compatibility to 1.8 (Java 8) Screenshot:
In this particular example, I think @Tagir is 100% correct get it into one filter and do the two checks. I wouldn’t use Optional.ofNullable the Optional stuff is really for return types not to be doing logic… but really neither here nor there. I wanted to point out that java.util.Objects has a nice method for … Read more
This will print 1 to 5. The return@forEach acts like the keyword continue in Java, which means in this case, it still executes every loop but skips to the next iteration if the value is greater than 5. fun main(args: Array<String>) { val nums = listOf(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) … Read more
You can adapt the solution you linked in How to sort ArrayList<Long> in Java in decreasing order? by wrapping it in a lambda: .sorted((f1, f2) -> Long.compare(f2.lastModified(), f1.lastModified()) note that f2 is the first argument of Long.compare, not the second, so the result will be reversed.
lambda is an anonymous function, it is equivalent to: def func(p): return p.totalScore Now max becomes: max(players, key=func) But as def statements are compound statements they can’t be used where an expression is required, that’s why sometimes lambda‘s are used. Note that lambda is equivalent to what you’d put in a return statement of a … Read more
You can use the ThenBy and ThenByDescending extension methods: foobarList.OrderBy(x => x.Foo).ThenBy( x => x.Bar)