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The internal representations of bit-field structures have several properties (such as internal padding) that are implementation-defined. Additionally, bit-field structures have several implementation-defined constraints:

• The alignment of bit fields in the storage unit. For example, the bit fields may be allocated from the high end or the low end of the storage unit.
• Whether or not bit fields can overlap a storage unit boundary.

Consequently, it is impossible to write portable safe code that makes assumptions regarding the layout of bit-field structure members.

Noncompliant Code Example (Bit-Field Alignment)

Bit fields can be used to allow flags or other integer values with small ranges to be packed together to save storage space. Bit fields can improve the storage efficiency of structures. Compilers typically allocate consecutive bit-field structure members into the same int-sized storage, as long as they fit completely into that storage unit. However, the order of allocation within a storage unit is implementation-defined. Some implementations are right-to-left: the first member occupies the low-order position of the storage unit. Others are left-to-right: the first member occupies the high-order position of the storage unit. Calculations that depend on the order of bits within a storage unit may produce different results on different implementations.

Consider the following structure made up of four 8-bit bit-field members:

struct bf {
  unsigned int m1 : 8;
  unsigned int m2 : 8;
  unsigned int m3 : 8;
  unsigned int m4 : 8;
};	/* 32 bits total */

Right-to-left implementations will allocate struct bf as one storage unit with this format:

m4   m3   m2   m1

Conversely, left-to-right implementations will allocate struct bf as one storage unit with this format:

m1   m2   m3   m4

The following code behaves differently depending on whether the implementation is left-to-right or right-to-left:

struct bf {
  unsigned int m1 : 8;
  unsigned int m2 : 8;
  unsigned int m3 : 8;
  unsigned int m4 : 8;
}; /* 32 bits total */

void function() {
  struct bf data;
  unsigned char *ptr;

  data.m1 = 0;
  data.m2 = 0;
  data.m3 = 0;
  data.m4 = 0;
  ptr = (unsigned char *)&data;
  (*ptr)++; /* can increment data.m1 or data.m4 */
}

Compliant Solution (Bit-Field Alignment)

This compliant solution is explicit in which fields it modifies.

struct bf {
  unsigned int m1 : 8;
  unsigned int m2 : 8;
  unsigned int m3 : 8;
  unsigned int m4 : 8;
}; /* 32 bits total */

void function() {
  struct bf data;
  data.m1 = 0;
  data.m2 = 0;
  data.m3 = 0;
  data.m4 = 0;
  data.m1++;
}

Noncompliant Code Example (Bit-Field Overlap)

In the following noncompliant code, assuming 8 bits to a byte, if bit fields of 6 and 4 bits are declared, is each bit field contained within a byte, or are the bit fields split across multiple bytes?

struct bf {
  unsigned int m1 : 6;
  unsigned int m2 : 4;
};

void function() {
  unsigned char *ptr;
  struct bf data;
  data.m1 = 0;
  data.m2 = 0;
  ptr = (unsigned char *)&data;
  ptr++;
  *ptr += 1; /* what does this increment? */
}

If each bit field lives within its own byte, then m2 (or m1, depending on alignment) is incremented by 1. If the bit fields are indeed packed across 8-bit bytes, then m2 might be incremented by 4.

Compliant Solution (Bit-Field Overlap)

This compliant solution is explicit in which fields it modifies.

struct bf {
  unsigned int m1 : 6;
  unsigned int m2 : 4;
};

void function() {
  struct bf data;
  data.m1 = 0;
  data.m2 = 0;
  data.m2 += 1;
}

Risk Assessment

Making invalid assumptions about the type of type-cast data, especially bit fields, can result in unexpected data values.

Automated Detection

Related Vulnerabilities

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

Related Guidelines

Bibliography