与AI合作 -- 单例工厂2遗留的问题：bard的错误

问题

上一节我们针对函数内静态变量初始化在多线程环境中要不要用锁保护提出了疑问，代码如下：

class Singleton {
public:
    static Singleton& getInstance() {
        std::lock_guard lock(mutex); // Acquire lock for thread safety
        static Singleton instance; // Local static variable for thread safety
        return instance;

结论

不必。C++11已经自己加锁了。

__cxa_guard_acquire

为了简单，我们先简单写个试验程序：

class Product {
public:
        Product(){
                std::cout<<"Product constructor"<

直接用GDB或者objdump看下test函数对应的汇编：

(gdb) disass test
Dump of assembler code for function test():
   0x0000000000400e26 <+0>:     push   %rbp
   0x0000000000400e27 <+1>:     mov    %rsp,%rbp
   0x0000000000400e2a <+4>:     push   %r12
   0x0000000000400e2c <+6>:     push   %rbx
   0x0000000000400e2d <+7>:     mov    $0x401808,%esi
   0x0000000000400e32 <+12>:    mov    $0x6030c0,%edi
   0x0000000000400e37 <+17>:    callq  0x400ca0 <_ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc@plt>
   0x0000000000400e3c <+22>:    mov    $0x400c40,%esi
   0x0000000000400e41 <+27>:    mov    %rax,%rdi
   0x0000000000400e44 <+30>:    callq  0x400cd0 <_ZNSolsEPFRSoS_E@plt>
   0x0000000000400e49 <+35>:    movzbl 0x202390(%rip),%eax        # 0x6031e0 <_ZGVZ4testvE3pro>
   0x0000000000400e50 <+42>:    test   %al,%al
   0x0000000000400e52 <+44>:    sete   %al
   0x0000000000400e55 <+47>:    test   %al,%al
   0x0000000000400e57 <+49>:    je     0x400e86 
   0x0000000000400e59 <+51>:    mov    $0x6031e0,%edi
   0x0000000000400e5e <+56>:    callq  0x400d20 <__cxa_guard_acquire@plt>
   0x0000000000400e63 <+61>:    test   %eax,%eax
   0x0000000000400e65 <+63>:    setne  %al
   0x0000000000400e68 <+66>:    test   %al,%al
   0x0000000000400e6a <+68>:    je     0x400e86 
   0x0000000000400e6c <+70>:    mov    $0x0,%r12d
   0x0000000000400e72 <+76>:    mov    $0x6031d9,%edi
   0x0000000000400e77 <+81>:    callq  0x40103a 
   0x0000000000400e7c <+86>:    mov    $0x6031e0,%edi
   0x0000000000400e81 <+91>:    callq  0x400c70 <__cxa_guard_release@plt>
   0x0000000000400e86 <+96>:    mov    $0x401817,%esi
   0x0000000000400e8b <+101>:   mov    $0x6030c0,%edi
   0x0000000000400e90 <+106>:   callq  0x400ca0 <_ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc@plt>
   0x0000000000400e95 <+111>:   mov    $0x400c40,%esi
   0x0000000000400e9a <+116>:   mov    %rax,%rdi
   0x0000000000400e9d <+119>:   callq  0x400cd0 <_ZNSolsEPFRSoS_E@plt>
   0x0000000000400ea2 <+124>:   jmp    0x400ec1 
   0x0000000000400ea4 <+126>:   mov    %rax,%rbx
   0x0000000000400ea7 <+129>:   test   %r12b,%r12b
   0x0000000000400eaa <+132>:   jne    0x400eb6 
   0x0000000000400eac <+134>:   mov    $0x6031e0,%edi
   0x0000000000400eb1 <+139>:   callq  0x400c60 <__cxa_guard_abort@plt>
   0x0000000000400eb6 <+144>:   mov    %rbx,%rax
   0x0000000000400eb9 <+147>:   mov    %rax,%rdi
   0x0000000000400ebc <+150>:   callq  0x400d10 <_Unwind_Resume@plt>
   0x0000000000400ec1 <+155>:   pop    %rbx
   0x0000000000400ec2 <+156>:   pop    %r12
   0x0000000000400ec4 <+158>:   pop    %rbp
   0x0000000000400ec5 <+159>:   retq

简单提一下_ZGVZ4testvE3pro demangle一下（第一次值是0，初始化后是1）：

[mzhai@singletonFactoryAi]$ c++filt _ZGVZ4testvE3pro
guard variable for test()::pro

 着重看下面的__cxa_guard_acquire

看到__cxa_guard_acquire、__cxa_guard_release没？中间夹着Product::Product()

即使没见过这两个函数，也能猜到和static变量初始化有关，防止多个线程同时初始化一个静态变量. 这是我把__cxa_guard_acquire输入搜索引擎搜到的：

C++ constructor guards for static instances

Introduction

The libstd++ ABI requires implementations of the following functions:

int __cxa_guard_acquire(__guard *g);

void __cxa_guard_release(__guard *g);

void __cxa_guard_abort(__guard *g);

These functions are not intended to be ever used by the programmer, instead the C++ compiler will emit code calling them if statically allocated class instances are used. In a multi-threaded environment special care needs to be taken to prevent race conditions while initializing and using such instances. This modules provides them.

Usage

This module is intended to be used by platforms that want to provide C++ support, but the used standard C++ library does not provide these guards. In this case, adding this module will do the trick. The programmer / user should never interact with any of the functions.

Note that on some platforms the type __guard is defined differently from the "generic" definition, most notably ARM. For those platforms a header named cxx_ctor_guards_arch.h needs to be created containing the correct typedef and the preprocessor macro CXX_CTOR_GUARDS_CUSTOM_TYPE needs to be defined.

https://opensource.apple.com/source/libcppabi/libcppabi-14/src/cxa_guard.cxx 

int __cxxabiv1::__cxa_guard_acquire(uint64_t* guard_object)
{
    // Double check that the initializer has not already been run
    if ( initializerHasRun(guard_object) )
        return 0;

    // We now need to acquire a lock that allows only one thread
    // to run the initializer.  If a different thread calls
    // __cxa_guard_acquire() with the same guard object, we want 
    // that thread to block until this thread is done running the 
    // initializer and calls __cxa_guard_release().  But if the same
    // thread calls __cxa_guard_acquire() with the same guard object,
    // we want to abort.  
    // To implement this we have one global pthread recursive mutex 
    // shared by all guard objects, but only one at a time.  

用GDB 模拟race condition

如果读者感兴趣，可以调试一下我们给出的程序：

(gdb) b __cxa_guard_acquire       使得两个线程都停在__cxa_guard_acquire
Breakpoint 2 at 0x7ffff7acf980
(gdb) r
Continuing.
Thread 2 "a.out" hit Breakpoint 2, 0x00007ffff7acf980 in __cxa_guard_acquire () from /lib64/libstdc++.so.6

Thread 3 "a.out" hit Breakpoint 2, 0x00007ffff7acf980 in __cxa_guard_acquire () from /lib64/libstdc++.so.6
Quit

(gdb) where
Selected thread is running.

(gdb) info thread                2、3线程都停在了__cxa_guard_acquire
  Id   Target Id                                   Frame
* 1    Thread 0x7ffff7fe1740 (LWP 2468153) "a.out" (running)
  2    Thread 0x7ffff6eb9700 (LWP 2468165) "a.out" 0x00007ffff7acf980 in __cxa_guard_acquire () from /lib64/libstdc++.so.6
  3    Thread 0x7ffff66b8700 (LWP 2468166) "a.out" 0x00007ffff7acf980 in __cxa_guard_acquire () from /lib64/libstdc++.so.6

(gdb) b Product::Product thread 2       
Breakpoint 3 at 0x401046: file static_var_in_func.cpp, line 36.
(gdb) thread apply 2 c            让第二个线程获得锁，先停在静态变量初始化中
Thread 2 (Thread 0x7ffff6eb9700 (LWP 2468165)):
Continuing.
Thread 2 "a.out" hit Breakpoint 3, Product::Product (this=0x6031d9 ) at static_var_in_func.cpp:36
36                      std::cout<<"Product constructor"< (__f=@0x616008: 0x400e26 ) at /usr/include/c++/8/bits/invoke.h:60
#4  0x0000000000401085 in std::__invoke (__fn=@0x616008: 0x400e26 ) at /usr/include/c++/8/bits/invoke.h:95
#5  0x00000000004016d4 in std::thread::_Invoker >::_M_invoke<0ul> (this=0x616008) at /usr/include/c++/8/thread:244
#6  0x00000000004016aa in std::thread::_Invoker >::operator() (this=0x616008) at /usr/include/c++/8/thread:253
#7  0x000000000040168e in std::thread::_State_impl > >::_M_run (this=0x616000) at /usr/include/c++/8/thread:196
#8  0x00007ffff7afbb23 in execute_native_thread_routine () from /lib64/libstdc++.so.6
#9  0x00007ffff78211ca in start_thread () from /lib64/libpthread.so.0
#10 0x00007ffff7275e73 in clone () from /lib64/libc.so.6

AI的回答

mzhai: Assume there is a static variable A which is an object of a class, this static variable is defined in a function named funA. When is A initialized? and is there a race condition if two threads call funA concurrently?

ChatGPT: In C++, static variables inside functions (local static variables) are initialized the first time the program execution encounters their declaration within that function. They're initialized in a thread-safe manner in compliance with the C++11 standard.

For instance, if you have a function funA with a static variable A inside it, like this:

void funA() { static MyClass A; // ... } 

When funA is called for the first time, the static MyClass A; line will be executed. This initialization occurs once, and subsequent calls to funA won't reinitialize A. This initialization is thread-safe in C++11 and later, meaning the standard guarantees that it will be properly handled in a multi-threaded context. There won't be a race condition during this initialization.

Bard: 

Here's a breakdown of the initialization and potential race condition:

1. Initialization of static variable A:

• When: The static variable A will be initialized the first time control flow reaches its declaration within funA, even if it's not explicitly used in that particular call.
• How: The constructor of the class will be invoked to initialize A's members.
• Subsequent calls: In subsequent calls to funA, the already initialized instance of A will be used, and the constructor won't be called again.

2. Race condition in multi-threaded environment:

• Yes, a race condition can occur if two threads call funA concurrently before A has been initialized.
• Scenario:
  • Both threads enter funA for the first time.
  • Both threads see that A hasn't been initialized yet.
  • Both threads attempt to create an instance of A, leading to multiple initializations and potential data corruption or crashes.

 可见在这方面Bard不如Chat GPT正确。

他山之石

后来搜到一篇外文博客，读者可以作为参考。Adventures in Systems Programming: C++ Local Statics - In Pursuit of Laziness