编译器在递归程序中的优化

Question

I got motivated from tail call optimization question What Is Tail Call Optimization?

So, I decided to see how can I do it in plain C.

So, I wrote the 2 factorial programs, 1st where tail call optimization can be applied. I call this fact function as fact(n, 1).

unsigned long long int fact(int n, int cont)
{
      if(n == 0)
            return cont;

      else return fact(n-1, n * cont);
}

2nd is the normal recursion where multiple stack frames are required.

unsigned long long int fact(int n)
{
    if(n == 0)
        return 1;

    else return n * fact(n-1);
}

This is the assembly generated by a 32 bit compiler for the former with -O2

0x8048470 <fact>:   push   %ebp
0x8048471 <fact+1>: mov    %esp,%ebp
0x8048473 <fact+3>: mov    0x8(%ebp),%edx
0x8048476 <fact+6>: mov    0xc(%ebp),%eax
0x8048479 <fact+9>: test   %edx,%edx
0x804847b <fact+11>:    je     0x8048488 <fact+24>
0x804847d <fact+13>:    lea    0x0(%esi),%esi
0x8048480 <fact+16>:    imul   %edx,%eax
0x8048483 <fact+19>:    sub    $0x1,%edx
0x8048486 <fact+22>:    jne    0x8048480 <fact+16>
0x8048488 <fact+24>:    mov    %eax,%edx
0x804848a <fact+26>:    sar    $0x1f,%edx
0x804848d <fact+29>:    pop    %ebp
0x804848e <fact+30>:    ret    

This is the assembly generated by a 32 bit compiler for latter with -O2.

0x8048470 <fact>:   push   %ebp
0x8048471 <fact+1>: mov    %esp,%ebp
0x8048473 <fact+3>: push   %edi
0x8048474 <fact+4>: push   %esi
0x8048475 <fact+5>: push   %ebx
0x8048476 <fact+6>: sub    $0x14,%esp
0x8048479 <fact+9>: mov    0x8(%ebp),%eax
0x804847c <fact+12>:    movl   $0x1,-0x18(%ebp)
0x8048483 <fact+19>:    movl   $0x0,-0x14(%ebp)
0x804848a <fact+26>:    test   %eax,%eax
0x804848c <fact+28>:    je     0x80484fc <fact+140>
0x804848e <fact+30>:    mov    %eax,%ecx
0x8048490 <fact+32>:    mov    %eax,%esi
0x8048492 <fact+34>:    sar    $0x1f,%ecx
0x8048495 <fact+37>:    add    $0xffffffff,%esi
0x8048498 <fact+40>:    mov    %ecx,%edi
0x804849a <fact+42>:    mov    %eax,%edx
0x804849c <fact+44>:    adc    $0xffffffff,%edi
0x804849f <fact+47>:    sub    $0x1,%eax
0x80484a2 <fact+50>:    mov    %eax,-0x18(%ebp)
0x80484a5 <fact+53>:    movl   $0x0,-0x14(%ebp)
0x80484ac <fact+60>:    sub    -0x18(%ebp),%esi
0x80484af <fact+63>:    mov    %edx,-0x20(%ebp)
0x80484b2 <fact+66>:    sbb    -0x14(%ebp),%edi
0x80484b5 <fact+69>:    movl   $0x1,-0x18(%ebp)
0x80484bc <fact+76>:    movl   $0x0,-0x14(%ebp)
0x80484c3 <fact+83>:    mov    %ecx,-0x1c(%ebp)
0x80484c6 <fact+86>:    xchg   %ax,%ax
0x80484c8 <fact+88>:    mov    -0x14(%ebp),%ecx
0x80484cb <fact+91>:    mov    -0x18(%ebp),%ebx
0x80484ce <fact+94>:    imul   -0x1c(%ebp),%ebx
0x80484d2 <fact+98>:    imul   -0x20(%ebp),%ecx
0x80484d6 <fact+102>:   mov    -0x18(%ebp),%eax
0x80484d9 <fact+105>:   mull   -0x20(%ebp)
0x80484dc <fact+108>:   add    %ebx,%ecx
0x80484de <fact+110>:   add    %ecx,%edx
0x80484e0 <fact+112>:   addl   $0xffffffff,-0x20(%ebp)
0x80484e4 <fact+116>:   adcl   $0xffffffff,-0x1c(%ebp)
0x80484e8 <fact+120>:   mov    -0x1c(%ebp),%ebx
0x80484eb <fact+123>:   mov    %eax,-0x18(%ebp)
0x80484ee <fact+126>:   mov    -0x20(%ebp),%eax
0x80484f1 <fact+129>:   mov    %edx,-0x14(%ebp)
0x80484f4 <fact+132>:   xor    %edi,%ebx
0x80484f6 <fact+134>:   xor    %esi,%eax
0x80484f8 <fact+136>:   or     %eax,%ebx
0x80484fa <fact+138>:   jne    0x80484c8 <fact+88>
0x80484fc <fact+140>:   mov    -0x18(%ebp),%eax
0x80484ff <fact+143>:   mov    -0x14(%ebp),%edx
0x8048502 <fact+146>:   add    $0x14,%esp
0x8048505 <fact+149>:   pop    %ebx
0x8048506 <fact+150>:   pop    %esi
0x8048507 <fact+151>:   pop    %edi
0x8048508 <fact+152>:   pop    %ebp
0x8048509 <fact+153>:   ret    

Compiling both the programs and looking at the assembly generated, both the programs still have recursive calls. But, when I compile with -O2 option(assembly posted above) in the former, I see no recursive calls at all and so I think gcc does tail call optimization stuff.

But when I compile the latter with -O2 option, it also removes the recursive calls and instead puts quite a large number of assembly instructions as compared to what happens to former on -O2.

I wanted to understand precisely what is the compiler doing in the latter, and why couldnt it transform to the assembly generated by former even with O4.

Answer 1

程序2进行long long计算，而progtlram 1则没有。

Answer 2

区别在于第一个版本使用int变量进行计算，然后在结束时将其扩展为unsigned long long ，而后者一直使用unsigned long long 。

Answer 3

The compiler appears to have optimised the recursive calls into loops. Note that your C code only has forward branches (if-then-else), but the assembler has backward branches (loops).

If you really want to see a tail-call optimization in action, have it call a different function. Of course, that's not recursion, but the compiler is too smart for small test-cases like this.