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:
Consequently, it is impossible to write portable safe code that makes assumptions regarding the layout of bit-field structure members.
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 */
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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 */
}
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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++;
}
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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? */
}
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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.
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;
}
|
Making invalid assumptions about the type of type-cast data, especially bit-fields, can result in unexpected data values.
Recommendation | Severity | Likelihood | Remediation Cost | Priority | Level |
|---|---|---|---|---|---|
EXP11-C | Medium | Probable | Medium | P8 | L2 |
Tool | Version | Checker | Description |
|---|---|---|---|
|
| Can detect violations of this recommendation. Specifically, it reports violations if
| |
94 S | Fully implemented | ||
| PRQA QA-C | 0310 | Partially implemented |
Search for vulnerabilities resulting from the violation of this recommendation on the CERT website.
| SEI CERT C++ Coding Standard | EXP11-CPP. Do not apply operators expecting one type to data of an incompatible type |
| ISO/IEC TR 24772:2013 | Bit Representations [STR] |
| MISRA C:2012 | Directive 1.1 (required) |
| [Plum 1985] | Rule 6-5: In portable code, do not depend upon the allocation order of bit-fields within a word |