gcc/g++ 中的尾递归 [英] Tail recursion in gcc/g++

问题描述

我尝试搜索，但找不到:函数的必要条件是什么，以便 gcc 优化尾递归?是否有任何参考或列表包含最重要的案例?由于这是特定于版本的，我感兴趣的是 4.6.3 或更高版本(越新越好).然而，事实上，任何具体的参考将不胜感激.

I've tried to search, but was unable to find: what are requisites for functions so that gcc would optimize the tail recursion? Is there any reference or list that would contain the most important cases? Since this is version specific, of my interests are 4.6.3 or higher versions (the newer the better). However, in fact, any concrete reference would be greatly appreciated.

提前致谢！

推荐答案

启用 -O2 后，如果可能，gcc 将执行尾调用优化.现在的问题是:什么时候可能?

With -O2 enabled, gcc will perform tail call optimization, if possible. Now, the question is: When is it possible?

只要递归调用发生在(单个)return 语句之前或内部，就可以消除单个递归调用，因此没有可观察到的副作用之后可能会返回不同的值.这包括返回涉及三元运算符的表达式.
请注意，即使返回表达式中有几次递归调用(例如在著名的斐波那契示例中:return (n==1) ? 1 : fib(n-1)+fib(n-2);)，尾调用递归仍然适用，但只适用于最后一次递归调用.

It is possible to eliminate a single recursive call whenever the recursive call happens either immediately before or inside the (single) return statement, so there are no observable side effects afterwards other than possibly returning a different value. This includes returning expressions involving the ternary operator.
Note that even if there are several recursive calls in the return expression (such as in the well-known fibonacci example: return (n==1) ? 1 : fib(n-1)+fib(n-2);), tail call recursion is still applied, but only ever to the last recursive call.

作为一个明显的特殊情况，如果编译器可以证明递归函数(及其参数)符合常量表达式的条件，它可以(最多可配置的最大递归深度，默认为 500)不仅消除尾调用，而是整个函数的执行.这是 GCC 在 -O2 上例行且相当可靠的事情，其中​​甚至包括对某些库函数的调用，例如文字上的 strlen.

As an obvious special case, if the compiler can prove that the recursive function (and its arguments) qualifies as constant expression, it can (up to a configurable maximum recursion depth, by default 500) eliminate not only the tail call, but the entire execution of the function. This is something GCC does routinely and quite reliably at -O2, which even includes calls to certain library functions such as strlen on literals.

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