lang无限尾递归优化

Question

#include <iostream> 

int foo(int i){ 
     return foo(i + 1);
} 

int main(int argc,char * argv[]){ 
     if(argc != 2){ 
         return 1; 
     } 
     std::cout << foo(std::atoi(argv[1])) << std::endl; 
} 

% clang++ -O2 test.cc

% time ./a.out 42

1490723512

./a.out 42 0.00s user 0.00s system 69% cpu 0.004 total

% time ./a.out 42

1564058296

./a.out 42 0.00s user 0.00s system 56% cpu 0.006 total

% g++ -O2 test.cc

% ./a.out 42 #infinte recursion

^C

% clang++ --version 
clang version 3.3 (tags/RELEASE_33/final) 
Target: x86_64-apple-darwin12.4.0 
Thread model: posix 
% g++ --version 
i686-apple-darwin11-llvm-g++-4.2 (GCC) 4.2.1 (Based on Apple Inc. build 5658) (LLVM build 2336.11.00) 
Copyright (C) 2007 Free Software Foundation, Inc. 
This is free software; see the source for copying conditions.  There is NO 
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. 

So is it a bug or a feature of clang++?

Answer 1

While both g++ and clang++ are able to compile C++98 and C++11 code, clang++ was designed from the start as a C++11 compiler and has some C++11 behaviors embedded in its DNA (so to speak).

With C++11 the C++ standard became thread aware, and that means that now there are some specific thread behavior. In particular 1.10/24 states:

The implementation may assume that any thread will eventually do one of the following:

— terminate,

— make a call to a library I/O function,

— access or modify a volatile object, or

— perform a synchronization operation or an atomic operation.

[Note: This is intended to allow compiler transformations such as removal of empty loops, even when termination cannot be proven. — end note ]

And that is precisely what clang++ is doing when optimizing. It sees that the function has no side effects and removes it even if it does not terminate.