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For tail calls, there is no need to remember the caller – instead, tail-call elimination makes only the minimum necessary changes to the stack frame before passing it on, [4] and the tail-called function will return directly to the original caller. The tail call doesn't have to appear lexically after all other statements in the source code ...
Remove the unused allocation and free. All of these steps are individually possible. Even step four is possible despite the fact that functions like malloc and free have global side effects, since some compilers hardcode symbols such as malloc and free so that they can remove unused allocations from the code. [6]
Note that tail call optimization in general (when the function called is not the same as the original function, as in tail-recursive calls) may be more difficult to implement than the special case of tail-recursive call optimization, and thus efficient implementation of mutual tail recursion may be absent from languages that only optimize tail ...
Every call in CPS is a tail call, and the continuation is explicitly passed. Using CPS without tail call optimization (TCO) will cause not only the constructed continuation to potentially grow during recursion, but also the call stack. This is usually undesirable, but has been used in interesting ways—see the Chicken Scheme compiler. As CPS ...
A simple tail recursive parser can be written much like a recursive descent parser. The typical algorithm for parsing a grammar like this using an abstract syntax tree is: Parse the next level of the grammar and get its output tree, designate it the first tree, F
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Tail-call optimization A function call consumes stack space and involves some overhead related to parameter passing and flushing the instruction cache. Tail-recursive algorithms can be converted to iteration through a process called tail-recursion elimination or tail-call optimization. Deforestation (data structure fusion)
A compiler can use the results of escape analysis as a basis for optimizations: [1] Converting heap allocations to stack allocations. [2] If an object is allocated in a subroutine, and a pointer to the object never escapes, the object may be a candidate for stack allocation instead of heap allocation.