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The constraint on the language for this to work well is fairly natural and is actually described in the article: the language should follow lexical scoping. Un

• chrisseaton 8y ago
  You can have plain lexical scoping but still be unable to use a simple stack in all cases, due to closures. The author describes this additional constraint that they have on top of lexical scoping. > We have to be OK with only allocating new locals on the top of the stack, and we have to accept that we can only discard a local when nothing is above it on the stack.
  • Drup 8y ago
    Well, there are lot's of well known compiler techniques to handle closures in that context. It's a little bit out of scope of the article, but it's not very complicated either.
    • chrisseaton 8y ago
      I know that but you said the constraint needed was lexical scoping - that constraint is insufficient and you need additional constraints on the language design.
      • klmr 8y ago
        If you’re happy for the compiler to copy variables out of the closed-over scope when returning the closure, this can still be handled with a single stack. That’s what C++ lambdas do: “Closures” in C++ are locally-allocated structures that hold used variables as local (stack-allocated) member variables. Creating a closure copies closed-over variables (or pointers/references to them). Returning a closure from a function returns a logical copy (which can be optimised away) of the structure. (Don’t get me wrong, this obviously still implies additional constraints, but it gets fairly close to universal closures.)
        • chrisseaton 8y ago
          I don't really understand that point of view - you can use a single stack as long as you actually use the heap in addition to a single stack?
          • kccqzy 8y ago
            C++ closures do not, by themselves, use the heap. Every closure in C++ gets translated by the compiler to a unique type that contains either copies of or references to the objects being closed over. If you choose to use copies, then whether or not anything is allocated on the heap depends on the copy constructor; if you use references then there is no copying, but it's up to you to ensure lifetime.
            • jdmichal 8y ago
              This is roughly how Java works with anonymous types closing over variables also. That's why the variables must be declared `final`. It just copies the local values over into the anonymous type and calls it a day. Of course, since the only thing allocated on the stack are primitives and pointers, and everything on the heap is subject to garbage collection, this is a pretty straight-forward operation. I don't know if lambdas work the same way. I know in some ways they work like anonymous types, and not in others.
              • _old_dude_ 8y ago
                yes, it works the same way with lambdas, the lambda proxy (the class that implements the functional interface) contains the copy of the local values. Here is the code that generate the constructor of a lambda proxy http://hg.openjdk.java.net/jdk/jdk/file/3cabb47758c9/src/java.base/share/classes/java/lang/invoke/InnerClassLambdaMetafactory.java#l356 http://hg.openjdk.java.net/jdk/jdk/file/3cabb47758c9/src/jav...
          • klmr 8y ago
            C++ lambdas do not use the heap. As I wrote, they are stack allocated.