EXP11-A. Do not apply operators expecting one type to data of an incompatible type

C has very weak typing. It lets you type-cast memory to different types, allowing you to apply operations of one type to data of a different type. However, the internal representation of most types are system-dependent. Applying operations on improper types will likely yield non-portable code and produce unexpected results.

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.

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