MEM33-C. Use the correct syntax for flexible array members

Flexible array members are a special type of array where the last element of a structure with more than one named member has an incomplete array type; that is, the size of the array is not specified explicitly within the structure. A variety of different syntaxes have been used for declaring flexible array members. For C99-compliant implementations, use the syntax guaranteed valid by C99 [[ISO/IEC 9899:1999]]. Section 6.7.2.1, paragraph 16: "Structure and Union Specifiers", says:

In particular, the size of the structure is as if the flexible array member were omitted except that it may have more trailing padding than the omission would imply.

Noncompliant Code Example (Declaration)

In this noncompliant code, an array of size 1 is declared, but when the structure itself is instantiated, the size computed for malloc() is modified to account for the actual size of the dynamic array. This is the syntax used by ISO C89.

struct flexArrayStruct {
  int num;
  int data[1];
};

/* ... */

size_t array_size;
size_t i;

/* initialize array_size */

/* space is allocated for the struct */
struct flexArrayStruct *structP
  = (struct flexArrayStruct *)
     malloc(sizeof(struct flexArrayStruct)
          + sizeof(int) * (array_size - 1));
if (structP == NULL) {
  /* Handle malloc failure */
}
structP->num = 0;

/* access data[] as if it had been allocated
 * as data[array_size] */
for (i = 0; i < array_size; i++) {
  structP->data[i] = 1;
}

The problem with this code is that the only member that is guaranteed to be valid, by strict C99 definition, is structP->data[0]. Consequently, for all i > 0, the results of the assignment are undefined.

Implementation Details

The noncompliant example may be the only alternative for compilers that do not yet implement the C99 syntax.  Microsoft Visual Studio 2005 does not implement the C99 syntax.

Compliant Solution (Declaration)

This compliant solution uses the flexible array member to achieve a dynamically sized structure.

struct flexArrayStruct{
  int num;
  int data[];
};

/* ... */

size_t array_size;
size_t i;

/* Initialize array_size */

/* Space is allocated for the struct */
struct flexArrayStruct *structP = (struct flexArrayStruct *)
   malloc(sizeof(struct flexArrayStruct) + sizeof(int) * array_size);
if (structP == NULL) {
  /* Handle malloc failure */
}

structP->num = 0;

/* Access data[] as if it had been allocated
 * as data[array_size]
 */
for (i = 0; i < array_size; i++) {
  structP->data[i] = 1;
}

This compliant solution allows the structure to be treated as if it had declared the member data[] to be data[array_size] in a manner that conforms to the C99 standard.

However, some restrictions apply:

1. The incomplete array type must be the last element within the structure.
2. There cannot be an array of structures that contain flexible array members.
3. Structures that contain a flexible array member cannot be used as a member in the middle of another structure.
4. The sizeof operator cannot be applied to a flexible array.

Noncompliant Code Example (Declaration Example 2)

When using structures with a flexible array member you should never directly declare an instance of the structure.  The following code snippet illustrates this.

struct flexArrayStruct{
  int num;
  int data[];
};

/* ... */

struct flexArrayStruct flexStruct;
size_t array_size;
size_t i;

/* Initialize array_size */

/* Initialize structure */
flexStruct.num = 0;

for (i = 0; i < array_size; i++) {
  flexStruct.data[i] = 0;
}

The problem with this code is that the flexArrayStruct does not actually reserve space for the integer array data - it can't as the size hasn't been specified.  Thus, while initializing the num member to zero is ok, attempting to write even one value into data (i.e. data[0]) will likely overwrite memory not owned by the structure.

Compliant Code Example (Declaration Example 2)

The solution is to always declare pointers to structures containing a flexible array member and dynamically allocate memory for them.  The following code snippet illustrates this.

struct flexArrayStruct{
  int num;
  int data[];
};

/* ... */

struct flexArrayStruct *flexStruct;
size_t array_size;
size_t i;

/* Initialize array_size */

/* Dynamically allocate memory for the structure */
flexStruct = (struct flexArrayStruct *)
   malloc(sizeof(struct flexArrayStruct) + sizeof(int) * array_size);
if (flexStruct == NULL) {
  /* Handle malloc failure */
}

/* Initialize structure */
flexStruct->num = 0;

for (i = 0; i < array_size; i++) {
  flexStruct->data[i] = 0;
}

In this code snippet the resolves the issue by declaring a pointer to flexArrayStruct and then dynamically allocating memory for the pointer to point to.  In this case it is ok to access the elements of the data[] member as described in Section 6.7.2.1, paragraph 16.

Noncompliant Code Example (Reference)

In this noncompliant code, the flexible array structure is passed directly to a function which tries to print the array elements. This fails because passing the struct directly to the function actually makes a copy of the struct. When the struct is copied, the size of the flexible array member is not considered, consequently the array is truncated, and the function only has garbage in the array to print.

struct flexArrayStruct {
  int num;
  int data[1];
};

void print_array( struct flexArrayStruct structP) {
  size_t i;
  printf("Array is: ");
  for (i = 0; i < structP.num; i++) {
    printf("%d", structP.data[i]);
  }
  printf("\n");
}

size_t array_size;
size_t i;

/* initialize array_size */

/* space is allocated for the struct */
struct flexArrayStruct *structP
  = (struct flexArrayStruct *)
     malloc(sizeof(struct flexArrayStruct)
          + sizeof(int) * (array_size - 1));
if (structP == NULL) {
  /* Handle malloc failure */
}
structP->num = array_size;

for (i = 0; i < array_size; i++) {
  structP->data[i] = i;
}

print_array( *structP);

Compliant Solution (Reference)

Never allow a flexible array member to be copied. The above code can be fixed by changing the function to accept a pointer to the flexible array struct.

struct flexArrayStruct{
  int num;
  int data[];
};

void print_array( struct flexArrayStruct* structP) {
  size_t i;
  printf("Array is: ");
  for (i = 0; i < structP->num; i++) {
    printf("%d", structP->data[i]);
  }
  printf("\n");
}

/* ... */

size_t array_size = 10;
size_t i;

/* Initialize array_size */

/* Space is allocated for the struct */
struct flexArrayStruct *structP = (struct flexArrayStruct *)
  malloc(sizeof(struct flexArrayStruct) + sizeof(int) * array_size);
if (structP == NULL) {
  /* Handle malloc failure */
}

structP->num = array_size;

/* Access data[] as if it had been allocated
 * as data[array_size]
 */
for (i = 0; i < array_size; i++) {
  structP->data[i] = i;
}

print_array( structP);
return 0;

Risk Assessment

Failing to use the correct syntax can result in undefined behavior, although the incorrect syntax will work on most implementations.

Automated Detection

Compass/ROSE can detect some violations of this rule. In particular, it warns if the last element of a struct is an array with a small index (0 or 1).

Related Vulnerabilities

Search for vulnerabilities resulting from the violation of this rule on the CERT website.

References

[[ISO/IEC 9899:1999]] Section 6.7.2.1, "Structure and union specifiers"
[[McCluskey 01]] ;login:, July 2001, Volume 26, Number 4

MEM32-C. Detect and handle memory allocation errors      08. Memory Management (MEM)