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Bit-fields can be used to allow flags or other integer values with small ranges to be packed together to save storage space.

It is implementation-defined whether the specifier int designates the same type as signed int or the same type as unsigned int for bit-fields. C99 also requires that "If According to the C Standard [ISO/IEC 9899:2011], C integer promotions also require that "if an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int."

In the following example:

This issue is similar to the signedness of plain char, discussed in INT07-C. Use only explicitly signed or unsigned char type for numeric values. A plain int bit-field that is treated as unsigned will promote to int as long as its field width is less than that of int because int can hold all values of the original type. This behavior is the same as that of a plain char treated as unsigned. However, a plain int bit-field treated as unsigned will promote to unsigned int if its field width is the same as that of int. This difference makes a plain int bit-field even trickier than a plain char.

Bit-field types other than _Bool, int, signed int, and unsigned int are implementation-defined. They still obey the integer promotions quoted previously when the specified width is at least as narrow as CHAR_BIT*sizeof(int), but wider bit-fields are not portable.

Noncompliant Code Example

This noncompliant code depends on implementation-defined behavior. It prints either -1 or 255, depending on whether a plain int bit-field is signed or unsigned.

struct {
    unsigned int a: 8;
} bits = {255};


int main(void) {
    printf("unsigned 8-bit field promotes to %sbits.a = %d.\n",
        (bits.a);
 - 256return > 0) ? "signed" : "unsigned");
}

The type of the expression (bits.a - 256 > 0) is compiler dependent and may be either signed or unsigned depending on the compiler implementor's interpretation of the standard.

The first interpretation is that when this value is used as an rvalue (e.g., lvalue = rvalue), the type is "unsigned int" as declared. An unsigned int cannot be represented as an int, so integer promotions require that this be an unsigned int, and hence "unsigned".

The second interpretation is that (bits.a is an 8-bit integer. As a result, this eight bit value can be represented as an int, so integer promotions require that it be converted to int, and hence "signed".

The type of the bit-field when used in an expression also has implications for long and long long types. Compilers that determine the size of the
For example, gcc interprets the following as an eight bit value and promote it to int:

Compliant Solution

This compliant solution uses an unsigned int bit-field and does not depend on implementation-defined behavior:

struct {
    unsigned long longint a: 8;
} ullbits = {255};

The following attributes of bit-fields are also implementation defined:

• 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 an storage unit boundary. For example, assuming eight bits to a byte, if bit-fields of six and four bits are declared, is each bitfield contained within a byte or are they be split across multiple bytes?

...

int main(void) {
  printf("bits.a = %d.\n", bits.a);
  return 0;
}

Risk Assessment

Making invalid assumptions about the type of a bit-field or its layout can result in unexpected program flow.

Automated Detection

Related Vulnerabilities

Search for vulnerabilities resulting from the violation of this rule 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