Versions Compared

Key

  • This line was added.
  • This line was removed.
  • Formatting was changed.

...

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.   A similar problem is discussed in CON00-C. Avoid race conditions with multiple threads, but the issue described by this rule can be harder to  The problem is difficult to diagnose because it may not be obvious that the same memory location is being modified by multiple threads.

...

Code Block
bgColor#FFcccc
langc
struct multi_threaded_flags {
  unsigned int flag1 : 2;
  unsigned int flag2 : 2;
};

struct multi_threaded_flags flags;

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

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

The C Standard, 3.14, paragraph 3 [ISO/IEC 9899:2011], states:

...

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 (Bit-field,

...

C++11 and later, Mutex)

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

Code Block
bgColor#ccccff
langc
#include <threads.h><mutex>
 
struct multi_threaded_flags {
  unsigned int flag1 : 2;
  unsigned int flag2 : 2;
};

struct mtf_mutex {
  struct multi_threaded_flags s;
  mtx_tstd::mutex mutex;
};

struct mtf_mutex flags;

int thread1(void *arg) {
  if (thrd_success != mtx_lockstd::lock_guard<std::mutex> guard(&flags.mutex)) {
    /* Handle error */;
  }
  flags.s.flag1 = 1;
  if (thrd_success != mtx_unlock(&flags.mutex)) {
    /* Handle error */
  }
  return 0;
}
 
int thread2(void *arg) {
  if (thrd_success != mtx_lockstd::lock_guard<std::mutex> guard(&flags.mutex)) {
    /* Handle error */
  };
  flags.s.flag2 = 2;
  if (thrd_success != mtx_unlock(&flags.mutex)) {
    /* Handle error */
  }
  return 0;
}

Compliant Solution (

...

C++11)

In this compliant solution, two threads simultaneously modify two distinct non-bit-field members of a structure. Because the members occupy different bytes in memory, no concurrency protection is required.

Code Block
bgColor#ccccff
langc
struct multi_threaded_flags {
  unsigned char flag1;
  unsigned char flag2;
};
 
struct multi_threaded_flags flags;
 
int thread1(void *arg) {
  flags.flag1 = 1;
  return 0;
}

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

Unlike C99, C11 explicitly defines earlier versions of the standard, C++11 and later explicitly define a memory location and provides provide the following note in subclause 3.14.2 in [intro.memory] paragraph 4 [ISO/IEC 9899:201114882-2014]:

NOTE 1 Two threads of execution can update and access separate memory locations without interfering with each other.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 C99 or 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 C11C++11, there were no guarantees that these flags could be modified concurrently.

...

Rule

Severity

Likelihood

Remediation Cost

Priority

Level

CON32CON52-CCPP

Medium

Probable

Medium

P8

L2

Automated Detection

...

Search for vulnerabilities resulting from the violation of this rule on the CERT website.

Bibliography

[ISO/IEC 9899:201114882-2015]3.14, "Memory Location"[intro.memory]

 

...