Tail-recursion not happening

Question

I'm using g++ (Ubuntu 4.8.2-19ubuntu1) 4.8.2 in a C++ project. I wrote a function that kind of does this:

template<typename T, T (*funct)(int) >
multiset<T> Foo(const multiset<T>& bar, int iterations) {
    if (iterations == 0) return bar; 
    multiset<T> copy_bar(bar); 

    T baz = funct(copy_bar.size());

    if (baz.attr > 0)
        return Foo<T,funct>(copy_bar, 100);
    else 
        return Foo<T,funct>(bar, iterations - 1);    
}

I was getting bad_alloc() exception so I tested the function with gdb and turns out that there's no tail-recursion happening, which I was expecting since there are no statements after the return s.

NOTE : I tried with -O2 compilation flag but it didn't work

Answer 1

您的函数不是尾递归的，因为在递归调用之后还有很多工作要做：销毁copy_bar （具有非平凡的析构函数），也可能破坏baz （如果类型T也具有非平凡的析构函数）。

Answer 2

As indicated by @celtschk's answer , non-trivial destructors are preventing the compiler from treating the calls as truly tail-recursive. Even when your function is reduced to this:

template<typename T, T (*funct)(int) >
multiset<T> Foo(const multiset<T>& bar, int iterations) {
    if (iterations == 0) return bar;
    return Foo<T,funct>(bar, iterations - 1);
}

It is still not tail-recursive, because of the implicit calls to the non-trivial constructors and destructors of the result of the recursive function call.

Note, however, that the above function does become tail-recursive with a relatively minor change:

template<typename T, T (*funct)(int) >
const multiset<T>& Foo(const multiset<T>& bar, int iterations) {
    if (iterations == 0) return bar;
    return Foo<T,funct>(bar, iterations - 1);
}

Voila! The function now compiles down into a loop. How can we achieve this with your original code?

Unfortunately, it is a little tricky, because your code will sometimes return the copy, and sometimes the original argument. We have to correctly deal with both cases. The solution presented below is a variation of the idea David mentioned in his comments to his answer.

Assuming you want to keep your original Foo signature the same (and, since it sometimes returns the copy, there is no reason to believe you wouldn't want to leave the signature the same), we create a secondary version of Foo called Foo2 that returns a reference to a result object that is either the original bar parameter, or one local in your Foo . In addition, Foo creates place holder objects for the copy, a secondary copy (for switching), and one to hold the funct() call result:

template<typename T, T (*funct)(int) >
multiset<T> Foo(const multiset<T>& bar, int iterations) {
    multiset<T> copy_bar;
    multiset<T> copy_alt;
    T baz;
    return Foo2<T, funct>(bar, iterations, copy_bar, copy_alt, baz);
}

Since Foo2 will always end up returning a reference, this removes any issues with the recursive function result causing implicit construction and destruction.

Upon each iteration, if the copy is to be used as the next bar , we pass the copy in, but switch the order of the copy and alternate placeholders for the recursive call, so that the alternate is actually used to hold the copy on the next iteration. If the next iteration reuses bar as is, the order of the arguments don't change, the iterations counter is merely decremented.

template<typename T, T (*funct)(int) >
const multiset<T>& Foo2(
        const multiset<T>& bar, int iterations,
        multiset<T>& copy_bar, multiset<T>& copy_alt, T& baz) {
    if (iterations == 0) return bar;
    copy_bar = bar;
    baz = funct(copy_bar.size());
    if (baz.attr > 0) {
        return Foo2<T, funct>(copy_bar, 100, copy_alt, copy_bar, baz);
    } else {
        return Foo2<T, funct>(bar, --iterations, copy_bar, copy_alt, baz);
    }
}

Notice that only the original call to Foo pays the penalty of construction and destruction of the local objects, and Foo2 is now entirely tail-recursive.

Answer 3

I don't think @celschk is right, but rather @jxh in an answer that he wiped out, so lets revisit it.

The fact that there are local variables that need to be destroyed does not affect tail recursion in general. The optimization is turning recursion into a loop, and those variables can be internal to the loop and be destroyed in each pass.

The problem, I believe, stems from the fact that the argument is a reference and that depending on some condition each pass over the loop would have to refer to either an object outside of the function or the local copy inside this function. If you try to manually unroll the recursion into a loop you will find it quite hard to figure out what should the loop look like.

To transform the recursion into a loop you would have to create an additional variable outside of the loop to hold the value of the multimap , turn the reference into a pointer and update the pointer to one or the other object depending on the condition before jumping back to the beginning of the loop. The baz variable cannot be used for this (ie it cannot be pulled outside of the loop) as each pass makes a copy and I imagine some other transformations that you did not show in the code above, so you really need to create an additional variable. The compiler cannot create new variables for you.

At this point, I have to admit that yes, the issue here is that for one of the branches copy_var needs to be destroyed after the recursion completes (as a reference to it is passed down), so @celtschk is not 100% wrong... but he is when he points at baz as another potential reason for breaking the tail-recursion.