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Weak typing in C allows type casting memory to different types. Because the internal representation of most types is system dependent, applying operations intended for data of one type to data of a different type will likely yield non-portable code and produce unexpected results.

Noncompliant Code Example (Integers vs. Floating-Point Numbers)

The following noncompliant code demonstrates the perils of operating on data of incompatible types. An attempt is made to increment an integer type cast to a floating point type and a floating point cast to an integer type.

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);

The expected result is for both values to display as 1; however, 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 (Integers vs. Floating-Point Numbers)

In this compliant solution, the pointers are assigned to reference variables of compatible data types.

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);

On the same platform, this solution produces the expected output of:

int is 0, float is 0.000000
int is 1, float is 1.000000

Bit-Fields

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 (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)++; /* canCan 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 this noncompliant code example, assuming eight assuming 8 bits to a byte, if bit-fields of six and four 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; /* whatWhat 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

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Bibliography

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Image Added Image Added 9899:1999|AA. Bibliography#ISO/IEC 9899-1999]\] Section 6.7.2, "Type specifiers" \[[ISO/IEC PDTR 24772|AA. Bibliography#ISO/IEC PDTR 24772]\] "STR Bit Representations" \[[MISRA 2004|AA. Bibliography#MISRA 04]\] Rule 3.5 \[[Plum 1985|AA. Bibliography#Plum 85]\] Rule 6-5Image Removed      03. Expressions (EXP)      Image Modified