Question

I wanted to check whether g++ supports tail calling so I wrote this simple program to check it: http://ideone.com/hnXHv

using namespace std;

size_t st;

void PrintStackTop(const std::string &type)
{
    int stack_top;
    if(st == 0) st = (size_t) &stack_top;
    cout << "In " << type << " call version, the stack top is: " << (st - (size_t) &stack_top) << endl;
}

int TailCallFactorial(int n, int a = 1)
{
    PrintStackTop("tail");
    if(n < 2)
        return a;
    return TailCallFactorial(n - 1, n * a);
}

int NormalCallFactorial(int n)
{
    PrintStackTop("normal");
    if(n < 2)
        return 1;
    return NormalCallFactorial(n - 1) * n;
}


int main(int argc, char *argv[])
{
    st = 0;
    cout << TailCallFactorial(5) << endl;
    st = 0;
    cout << NormalCallFactorial(5) << endl;
    return 0;
}

When I compiled it normally it seems g++ doesn’t really notice any difference between the two versions:

> g++ main.cpp -o TailCall
> ./TailCall
In tail call version, the stack top is: 0
In tail call version, the stack top is: 48
In tail call version, the stack top is: 96
In tail call version, the stack top is: 144
In tail call version, the stack top is: 192
120
In normal call version, the stack top is: 0
In normal call version, the stack top is: 48
In normal call version, the stack top is: 96
In normal call version, the stack top is: 144
In normal call version, the stack top is: 192
120

The stack difference is 48 in both of them, while the tail call version needs one more
int. (Why?)
So I thought optimization might be handy:

> g++ -O2 main.cpp -o TailCall
> ./TailCall
In tail call version, the stack top is: 0
In tail call version, the stack top is: 80
In tail call version, the stack top is: 160
In tail call version, the stack top is: 240
In tail call version, the stack top is: 320
120
In normal call version, the stack top is: 0
In normal call version, the stack top is: 64
In normal call version, the stack top is: 128
In normal call version, the stack top is: 192
In normal call version, the stack top is: 256
120

The stack size increased in both cases, and while the compiler might think my CPU is slower than my memory (which its not anyway), I don’t know why 80 bytes are necessary for a simple function. (Why is it?).
There tail call version also takes more space than the normal version, and its completely logical if an int has the size of 16 bytes. (no, I don’t own a 128 bit CPU).
Now thinking what reason the compiler has not to tail call, I thought it might be exceptions, because they depend on the stack tightly. So I tried without exceptions:

> g++ -O2 -fno-exceptions main.cpp -o TailCall
> ./TailCall
In tail call version, the stack top is: 0
In tail call version, the stack top is: 64
In tail call version, the stack top is: 128
In tail call version, the stack top is: 192
In tail call version, the stack top is: 256
120
In normal call version, the stack top is: 0
In normal call version, the stack top is: 48
In normal call version, the stack top is: 96
In normal call version, the stack top is: 144
In normal call version, the stack top is: 192
120

Which cut the normal version back to non-optimized stack size, while the optimized one has 8 bytes over it. still an int is not 8 bytes.
I thought there is something I missed in c++ that needs the stack arranged so I tried c: http://ideone.com/tJPpc
Still no tail calling, but the stack is much smaller (32 bit each frame in both version).
Then I tried with optimization:

> gcc -O2 main.c -o TailCall
> ./TailCall
In tail call version, the stack top is: 0
In tail call version, the stack top is: 0
In tail call version, the stack top is: 0
In tail call version, the stack top is: 0
In tail call version, the stack top is: 0
120
In normal call version, the stack top is: 0
In normal call version, the stack top is: 0
In normal call version, the stack top is: 0
In normal call version, the stack top is: 0
In normal call version, the stack top is: 0
120

Not only it tail call optimized the first, it also tail call optimized the second!
Why doesn’t g++ do tail call optimization while its clearly available in the platform? is there any way to force it?

Answer 1

Because you’re passing a temporary std::string object to the PrintStackTop(std::string) function. This object is allocated on the stack and thus prevent the tail call optimization.

I modified your code:

void PrintStackTopStr(char const*const type)
{
    int stack_top;
    if(st == 0) st = (size_t) &stack_top;
    cout << "In " << type << " call version, the stack top is: " << (st - (size_t) &stack_top) << endl;
}

int RealTailCallFactorial(int n, int a = 1)
{
    PrintStackTopStr("tail");
    if(n < 2)
        return a;
    return RealTailCallFactorial(n - 1, n * a);
}

Compile with: g++ -O2 -fno-exceptions -o tailcall tailcall.cpp

And it now uses the tail call optimisation. You can see it in action if you use the -S flag to produce the assembly:

L39:
        imull   %ebx, %esi
        subl    $1, %ebx
L38:
        movl    $LC2, (%esp)
        call    __Z16PrintStackTopStrPKc
        cmpl    $1, %ebx
        jg      L39

You see the recursive call inlined as a loop (jg L39).