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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. When used in structure members, bit Bit-fields can improve the storage efficiency .

In portable code, do not depend upon the allocation order of bit-fields in memory. Of course, in machine-specific non-portable code one knows exactly how the bit-fields are laid out, and the internal details can be inspected with bitwise operations.

Consider the representation of time-of-day in hours, minutes, seconds and milliseconds. Bit-fields provide one way to represent such times:

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

typedef struct time_day {
  unsigned int h1m1 : 2;	{0:2}8;
  unsigned int m2 : 8;
  unsigned h2int 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  unsigned m2 m2   m3   m4

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

struct bf {
  unsigned s1int m1 : 8;
  unsigned s2 int m2 : 8;
  unsigned f1 int m3 : 8;
  unsigned f2 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;
};  f3
  {0:9} {0:5) {0:9) {0:5) {0:9) {0:9} {0:9} {0:9}
} TIME_DAY;	/* 32 bits total */

The last millisecond of the day is
23:59:59.999 (hh:mm:ss.fff)
Each member (bit-field) is declared to be unsigned (int); this is the only bit-field type that is guaranteed to be portable to all current compilers. Each member is declared to have only as many bits as are necessary to represent the possible digits at its position in the time representation. Representing h1 (first digit of hours) takes only two bits to represent the possible values (0, 1, and 2). And the largest members need only four bits to represent ten digits, 0 through 9. The total number of bits is 32.

Consecutive bit-field members are allocated by the compiler to the same int-sized word, as long as they fit completely. Consequently, on a 32-bit machine, a TIME_DAY object occupies exactly one {int}}-sized word. Such an exact fit is rare, however. Add another member such as "day-of-year" to the structure, and the nice size-fitting property disappears. Consequently, bit-fields are useful for storage-saving only if they occupy most or all of the space of an int, and if the storage-saving property is to be reasonably portable, they must occupy most of the space in a 32-bit integer.

The order of allocation within a word is different in different implementations. Some implementations are "right-to-left": the first member occupies the low-order position of the word. Following the convention that the low-order bit of a word is on the right, the right-to-left allocation would look like this:

f3	f2	f1	s2	s1	m2	ml	h2      h1

Most other implementations are "left-to-right":

h1	h2	ml	m2	s1	s2	f1	f2	f3

A union provides a convenient way to say what is going on:

typedef union time_overlay {	/* MACHINE DEPENDENT */
  struct time_day time_as_fields;
  long time_as_long;
} TIME_OVERLAY;

TIME_OVERLAY time_port;

This allows bitwise operations like time_port.time_as_long & 0xF00 as well as providing access via bit-field names like time_port.time_as_fields.h1.

Specifying a field size of zero causes any subsequent allocations to begin on a new word boundary. Un-named bit-fields are allowed; they occupy space but are inaccessible, which is useful for padding within a structure.

Because most C machines do not support bit-addressing, the "address-of" (&) operator is not allowed upon bit-field members.
Aside from these complications, bit-fields can be treated just like any other structure member. The following declaration

#include "time_day.h"
struct time_day last_msec = {2, 3, 5, 9, 5, 9, 9, 9, 9};
 /* initializes last_msec to the last millisecond of the day. */
struct time_day now;

/* ... */
if (now.h1 == 0 || (now.h1 == 1 && now.h2 < 2))

tests whether now is less than noon.

The TIME_DAY example illustrates the use of bit-fields nicely, but there are numerous other ways to represent time-of-day.

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 this noncompliant code example, 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 a type-cast data, especially bit-field or its layout fields, can result in unexpected program flowdata values.

Automated Detection

Related Vulnerabilities

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

References

Related Guidelines

Bibliography

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Image Added Image Added Image Added Wiki Markup\[[ISO/IEC 9899-1999|AA. C References#ISO/IEC 9899-1999]\] Section 6.7.2, "Type specifiers" \[[MISRA 04|AA. C References#MISRA 04]\] Rule 3.5, Rule 6.4, "Bit fields shall only be defined to be of type unsigned int or signed int" \[[Plum 85|AA. C References#Plum 85]\] Rule 6-5