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UNDER CONSTRUCTION

When accessing a bit-field, a thread may inadvertently access a separate bit-field in adjacent memory. This is because compilers are required to store multiple adjacent bit-fields in one storage unit whenever they fit. Consequently, data races may exist not just on a bit-field accessed by multiple threads but also on other bit-fields sharing the same byte or word. The The problem is difficult to diagnose because it may not be obvious that the same memory location is being modified by multiple threads.

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Another approach is to insert a non-bit-field member between any two bit-fields to ensure that each bit-field is the only one accessed within its storage unit. This technique effectively guarantees that no two bit-fields are accessed simultaneously.

Noncompliant Code Example (

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bit-field)

Adjacent bit-fields may be stored in a single memory location. Consequently, modifying adjacent bit-fields in different threads is undefined behavior, as shown in this noncompliant code example:.

Code Block

bgColor	#FFcccc
lang	c

struct multi_threaded_flagsMultiThreadedFlags {
  unsigned int flag1 : 2;
  unsigned int flag2 : 2;
};

struct multi_threaded_flagsMultiThreadedFlags flags;

intvoid thread1(void *arg) {
  flags.flag1 = 1;
  return 0;
}

intvoid thread2(void *arg) {
  flags.flag2 = 2;
  return 0;
}

For example, the following instruction sequence is possible:.

Code Block

Thread 1: register 0 = flags
Thread 1: register 0 &= ~mask(flag1)
Thread 2: register 0 = flags
Thread 2: register 0 &= ~mask(flag2)
Thread 1: register 0 |= 1 << shift(flag1)
Thread 1: flags = register 0
Thread 2: register 0 |= 2 << shift(flag2)
Thread 2: flags = register 0

Compliant Solution (

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bit-field, C++11 and later,

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mutex)

This compliant solution protects all accesses of the flags with a mutex, thereby preventing any data races:.

Code Block

bgColor	#ccccff
lang	c

#include <mutex>
 
struct multi_threaded_flagsMultiThreadedFlags {
  unsigned int flag1 : 2;
  unsigned int flag2 : 2;
};

struct mtf_mutexMtfMutex {
  struct multi_threaded_flagsMultiThreadedFlags s;
  std::mutex mutex;
};

struct mtf_mutexMtfMutex flags;

intvoid thread1(void *arg) {
  std::lock_guard<std::mutex> guardlk(&flags.mutex);
  flags.s.flag1 = 1;
  return 0;
}
 
intvoid thread2(void *arg) {
  std::lock_guard<std::mutex> guardlk(&flags.mutex);
  flags.s.flag2 = 2;
  return 0;
}

Compliant Solution (C++11)

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Code Block

bgColor	#ccccff
lang	c

struct multi_threaded_flagsMultiThreadedFlags {
  unsigned char flag1;
  unsigned char flag2;
};
 
struct multi_threaded_flagsMultiThreadedFlags flags;
 
intvoid thread1(void *arg) {
  flags.flag1 = 1;
  return 0;
}

intvoid thread2(void *arg) {
  flags.flag2 = 2;
  return 0;
}

Unlike earlier versions of the standard, C++11 and later explicitly define a memory location and provide the following note in [intro.memory] paragraph 4 [ISO/IEC 14882-2014]:

[ Note: Thus a bit-field and an adjacent non-bit-field are in separate memory locations, and therefore can be concurrently updated by two threads of execution without interference. The same applies to two bit-fields, if one is declared inside a nested struct declaration and the other is not, or if the two are separated by a zero-length bit-field declaration, or if they are separated by a non-bit-field declaration. It is not safe to concurrently update two bit-fields in the same struct if all fields between them are also bit-fields of non-zero width. – end note]

It is almost certain that flag1 and flag2 are stored in the same word. Using a compiler that conforms to earlier versions of the standard, if both assignments occur on a thread-scheduling interleaving that ends with both stores occurring after one another, it is possible that only one of the flags will be set as intended, and the other flag will contain its previous value , because both members are represented by the same word, which is the smallest unit the processor can work on. Before the changes made to the C++ Standard for C++11, there were no guarantees that these flags could be modified concurrently.

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Although the race window is narrow, an assignment or an expression can evaluate improperly because of misinterpreted data resulting in a corrupted running state or unintended information disclosure.