Recursion And Stack Overflow
Few days ago, the Twitter account of StackOverflow (the famous site of technical questions and answers) has published the following tweet:
The intention was to recall the name of the site by presenting a situation that every developer has been warned about.
Of course I got the pun but a light bulb has turned on in my mind: tail calls. Some years ago I've read something about it on the blog of Gustavo Duarte: Tail Calls, Optimization, and ES6.
Do It Yourself
It's not that I don't trust Gustavo, but I've tried to do my own test (on x86-64bits with gcc 4.8):
#include <stdio.h>
int main()
{
fprintf(stderr, "Go!\n");
return main();
}
Compiled with the optimization enabled:
$ gcc -O2 -o so so.c
Launching the executable so
(that stands for stack overflow), it prints a never ending series of "Go!", without crashing. The explanation is exactly the same Gustavo presented in his old post: the compiler optimizes this call with a jump. So basically it transforms a recursion in a loop. This is clear looking at the disassembled code, with the following command:
$ objdump -d so
This is the result (leaving only the interesting part):
0000000000400480 <main>:
400480: 48 83 ec 08 sub $0x8,%rsp
400484: 0f 1f 40 00 nopl 0x0(%rax)
400488: 48 8b 0d b1 0b 20 00 mov 0x200bb1(%rip),%rcx # 601040 <__TMC_END__>;
40048f: ba 04 00 00 00 mov $0x4,%edx
400494: be 01 00 00 00 mov $0x1,%esi
400499: bf 14 06 40 00 mov $0x400614,%edi
40049e: e8 cd ff ff ff callq 400470 <fwrite@plt>
4004a3: eb e3 jmp 400488 <main+0x8>
As you can see, fprintf
is mapped with a callq
to fwrite
, while the recursion is made by the instruction jmp
Conclusions
Basically there are three considerations that can be done:
- Gustavo was right;
- compilers are smarter than you think;
- the same thing I've wrote about goto and do-while loops apply also for recursion.
Cover image taken from Wikimedia Commons (public domain).