C has very weak typing. It lets you type-cast memory to different types, allowing you to apply operations of one type to apply to data of a different type. However, the internal representation of most types are system-dependent. Consequently, applying operations that expect a certain type of data will yield unexpected results when applied to data of the wrong type. Furthermore, applying operations on improper types will yield non-portable code, due to the platform-dependent representation of the data.
Non-Compliant Code Example (Ints vs. Floats)
The following code demonstrates the perils of operating on data of improper types. It tries to increment an int typecast as a float, and a float typecast as an int, and displays the results.
#include <assert.h> #include <stdio.h> int main() { float f = 0.0; int i = 0; float *fp; int *ip; assert(sizeof(int) == sizeof(float)); ip = (int*) &f; fp = (float*) &i; printf("int is %d, float is %f\n", i, f); (*ip)++; (*fp)++; printf("int is %d, float is %f\n", i, f); return 0; }
Rather than the int and float both having the value 1, on a 64-bit Linux machine, this program produces:
int is 0, float is 0.000000 int is 1065353216, float is 0.000000
Compliant Solution (Ints vs. Floats)
Here the pointers are assigned to the variables of the proper data types.
#include <stdio.h> int main() { float f = 0.0; int i = 0; float *fp; int *ip; ip = &i; fp = &f; printf("int is %d, float is %f\n", i, f); (*ip)++; (*fp)++; printf("int is %d, float is %f\n", i, f); return 0; }
This program, on the same platform, produces:
int is 0, float is 0.000000 int is 1, float is 1.000000
which is what one would expect.
Bit-Fields
The internal representation of bit-field structs have several properties that are implementation-defined. For instance, they may contain internal padding. Bit-field structures have several additional 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.
Non-Compliant 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)++; /* could 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++; }
Non-Compliant Code Example (Bit-Field Overlap)
In this non-compliant example, assuming eight bits to a byte, if bit-fields of six and four 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? */ }
In the above example, 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 code also violates ARR37-C. Do not add or subtract an integer to a pointer to a non-array object
Compliant Solution (Bit-Field Overlap)
This compliant solution is also 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 typecast data, especially bit-fields can result in unexpected data values.
Recommendation |
Severity |
Likelihood |
Remediation Cost |
Priority |
Level |
---|---|---|---|---|---|
INT11-A |
low |
unlikely |
medium |
P2 |
L3 |
Related Vulnerabilities
Search for vulnerabilities resulting from the violation of this rule on the CERT website.
References
[[ISO/IEC 9899-1999]] Section 6.7.2, "Type specifiers"
[[ISO/IEC PDTR 24772]] "STR Bit Representations"
[[MISRA 04]] Rule 3.5
[[Plum 85]] Rule 6-5
EXP10-A. Do not depend on the order of evaluation of subexpressions or the order in which side effects take place 03. Expressions (EXP) EXP30-C. Do not depend on order of evaluation between sequence points