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Wiki MarkupThe C99 standard \[[ISO/IEC 9899:1999|AA. C References#ISO/IEC 9899-1999]\] introduces flexible array members into the language. While flexible array members are a useful addition they should be properly understood and used with care. Flexible array members are defined in Section C Standard, 6.7.3.2.1, paragraph 16 of the C99 standard as follows,, paragraph 20 [ISO/IEC 9899:2024], says

As a special case, the last element of a structure with more than one named member may have an incomplete array type; this is called a flexible array member. In most situations, the flexible flexible array member is ignored. 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. However, when a . (or ->) operator has a left operand that is (a pointer to) a structure with a flexible array member and the right operand names that member, it behaves as if that member were replaced with the longest array (with the same element type) that would not make the structure larger than the object being accessed; the offset of the array shall remain that of the flexible array member, even if this would differ from that of the replacement array. If this array would have no elements, it behaves as if it had one element but the behavior is undefined if any attempt is made to access that element or to generate a pointer one past it.

The following is an example of a structure that contains a flexible array member,:

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

This definition means that , when allocating storage spacewhen computing the size of such a structure, only the first member, num, is considered. ConsequentlyUnless the appropriate size of the flexible array member has been explicitly added when allocating storage for an object of the struct, the result of accessing the member data of a variable of type struct flexArrayStruct nonpointer type struct flex_array_struct is undefined. DCL38-C. Use the correct syntax when declaring a flexible array member describes the correct way to declare a struct with a flexible array member.

To avoid the potential for undefined behavior, structures that contain a flexible array member should always be accessed with a pointer as shown in the following code example.

struct flexArrayStruct *structP;
size_t array_size;
size_t i;

/* Initialize array_size */

/* Space is allocated for the struct */
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;
}

Structures with flexible array members can be used to produce code with defined behavior.  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.

Noncompliant Code Example (Use Flexible Array Members)

When using C99 compliant compilers the one element array hack described above should not be used.  In this noncompliant code, just such an array is used where a flexible array member should be used instead.

allocated dynamically. Flexible array structures must

• Have dynamic storage duration (be allocated via malloc() or another dynamic allocation function)
• Be dynamically copied using memcpy() or a similar function and not by assignment
• When used as an argument to a function, be passed by pointer and not copied by value

Noncompliant Code Example (Storage Duration)

This noncompliant code example uses automatic storage for a structure containing a flexible array member:

#include <stddef.h>
 
struct flex_array_struct {
  size_t

struct flexArrayStruct {
  int num;
  int data[1];
};
 
/* ... */

struct flexArrayStruct *flexStruct;
void func(void) {
  struct flex_array_struct flex_struct;
  size_t array_size;
size_t i = 4;

/* Initialize array_size */

/* Dynamically allocate memory for theInitialize structure */
flexStruct = (flex_struct flexArrayStruct *)
   malloc(sizeof(struct flexArrayStruct) + sizeof(int) * (array_size - 1));
if (flexStruct == NULL) {
  /* Handle malloc failure */
}

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

for (.num = array_size;

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

As described above, the problem with this code is that strictly speaking the only member that is guaranteed to be valid is {{flexStruct->data\[0\]}}. Unfortunately, when using compilers that do not support the C99 standard in full, or at all, this approach may be the only solution.  Microsoft Visual Studio 2005, for example, does not implement the C99 syntax.

Compliant Solution (Use Flexible Array Members)

}

Because the memory for flex_struct is reserved on the stack, no space is reserved for the data member. Accessing the data member is undefined behavior.

Compliant Solution (Storage Duration)

This compliant solution dynamically allocates storage for flex_array_struct:Fortunately, when working with C99 compliant compilers, the solution is simple - remove the 1 from the array declaration and adjust the size passed to the malloc() call accordingly.  In other words, use flexible array members.

#include <stdlib.h>
 
struct flexArrayStructflex_array_struct {
  intsize_t num;
  int data[];
};
 
/* ... */

struct flexArrayStruct *flexStruct;
void func(void) {
  struct flex_array_struct *flex_struct;
  size_t array_size;
size_t i = 4;

/* Initialize array_size */

/* Dynamically allocate memory for the structurestruct */
flexStruct  flex_struct = (struct flexArrayStructflex_array_struct *)malloc(
    malloc(sizeof(struct flexArrayStructflex_array_struct)
    + sizeof(int) * array_size);
  if (flexStructflex_struct == NULL) {
    /* Handle mallocerror failure */
  }

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

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

...

...

...

}
}

Noncompliant Code Example (

...

Copying)

This noncompliant code example attempts to copy When using structures with a flexible array member you should never directly declare an instance of the structure.  In this noncompliant code, a variable of type struct flexArrayStruct is declared.a structure containing a flexible array member (struct flex_array_struct) by assignment:

#include <stddef.h>
 
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.  Consequently, while initializing the num member to zero is allowed, attempting to write even one value into data (that is, {{data\[0\]}}) will likely overwrite memory not owned by the structure.

Compliant Code Example (Declaration)

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 *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 acceptable to access the elements of the {{data\[\]}} member as described in C99 Section 6.7.2.1, paragraph 16.

Noncompliant Code Example (Copying)

When using structures with a flexible array member you should never directly copy an instance of the structure.  This noncompliant code attempts to replicate a copy of struct flexArrayStruct.

struct flexArrayStruct *flexStructA;
struct flexArrayStruct *flexStructB;
size_t array_size;
size_t i;

/* Initialize array_size */

/* Allocate memory for flexStructA */

/* Allocate memory for flexStructB */

/* Initialize flexStructA */

/* ... */

*flexStructB = *flexStructA;

flex_array_struct {
  size_t num;
  int data[];
};
 
void func(struct flex_array_struct *struct_a,
          struct flex_array_struct *struct_b) {
  *struct_b = *struct_a;
}

When the structure is copied, The problem with this code is that when the structure is copied the size of the flexible array member is not considered, and only the first member of the structure, num, is copied, leaving the array contents untouched.

Compliant Solution (Copying)

This compliant solution uses memcpy() to properly copy the content of flexStructA into flexStructB.of struct_a into struct_b:

#include <string.h>
 
struct flexArrayStruct *flexStructA;
struct flexArrayStruct *flexStructB;
flex_array_struct {
  size_t array_sizenum;
size_t i  int data[];
};
 
/* Initialize void func(struct flex_array_sizestruct */

/* Allocate memory for flexStructA */

/* Allocate memory for flexStructB */

/* Initialize flexStructA */

/* ... */

memcpy(flexStructB, flexStructA, (sizeof(struct flexArrayStruct) + sizeof(int) * array_size));

In this case the copy is explicit and the flexible array member is accounted for and copied as well.

Noncompliant Code Example (Reference)

struct_a,
          struct flex_array_struct *struct_b) {
  if (struct_a->num > struct_b->num) {
    /* Insufficient space; handle error */
    return;
  }
  memcpy(struct_b, struct_a,
         sizeof(struct flex_array_struct) + (sizeof(int)
           * struct_a->num));
}

Noncompliant Code Example (Function Arguments)

In this noncompliant code exampleWhen using structures with a flexible array member you should never directly pass an instance of the structure in a function call.  In this noncompliant code, the flexible array structure is passed directly by value to a function which tries to print that prints the array elements.:

void print_array(struct flexArrayStruct structP) #include <stdio.h>
#include <stdlib.h>
 
struct flex_array_struct {
  size_t i;
num;
  int data[];
};
 
void print_array(struct flex_array_struct struct_p) {
  printfputs("Array is: ");
  for (size_t i = 0; i < structPstruct_p.num; i++i) {
    printf("%d ", structPstruct_p.data[i]);
  }
  printfputchar("'\n"');
}

void func(void) {
  struct flexArrayStructflex_array_struct *structPstruct_p;
  size_t array_size;
size_t i = 4;

/* initialize array_size */

/* spaceSpace is allocated for the struct */
structP  struct_p = (struct flexArrayStructflex_array_struct *)malloc(
     malloc(sizeof(struct flexArrayStruct) flex_array_struct)
    + sizeof(int) * array_size);
  if (structPstruct_p == NULL) {
    /* Handle malloc failureerror */
  }

structP  struct_p->num = array_size;

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

  print_array(*structPstruct_p);
}

Because the argument is passed by value, the size of the flexible array member is not considered when the structure is copied, and only the first member of the structure, num, is copiedThe problem with this code is that passing the structure directly to the function actually makes a copy of the structure.  This copied fails for the same reason as the copy example above.

Compliant Solution (

...

Function Arguments)

Never allow a structure with a flexible array member to be passed directly in a function call.  The above code can be fixed by changing the function to accept a pointer to the structure.In this compliant solution, the structure is passed by reference and not by value:

#include <stdio.h>
#include <stdlib.h>
 
struct flex_array_struct
void print_array(struct flexArrayStruct *structP) {
  size_t inum;
  int data[];
};
 
void  printfprint_array(struct flex_array_struct *struct_p) {
  puts("Array is: ");
  for (size_t i = 0; i < structPstruct_p->num; i++i) {
    printf("%d ", structPstruct_p->data[i]);
  }
  printfputchar("'\n"');
}

void func(void) {
  struct flexArrayStructflex_array_struct *structPstruct_p;
  size_t array_size;
size_t i = 4;

/* initialize array_size */

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

structP->num = array_size;

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

print_array(structP);
and initialized... */

  print_array(struct_p);
}

Risk Assessment

Failure to use structures with flexible array members correctly can result in undefined behavior. 

Automated Detection

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Related Vulnerabilities

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

References

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Related Guidelines

Key here (explains table format and definitions)

CERT-CWE Mapping Notes

Key here for mapping notes

CWE-401 and MEM33-CPP

There is no longer a C++ rule for MEM33-CPP. (In fact, all C++ rules from 30-50 are gone, because we changed the numbering system to be 50-99 for C++ rules.)

Bibliography

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Image Added Image Added |http://www.open-std.org/jtc1/sc22/wg14/www/docs/n791.htm]\]MEM32-C. Detect and handle memory allocation errors      08. Memory Management (MEM)      Image Modified