Synopsis

Use the correct syntax when declaring a flexible array member.

Enabled by default

Yes

Severity/Certainty

Low/Low

Full description

A variety of different syntaxes have been used for declaring flexible array members. For conforming C implementations, use the syntax guaranteed to be valid by the C Standard.

Coding standards

Code examples

The following code example fails the check and will give a warning:

#include <stdlib.h>

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

void func(size_t 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;

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

The following code example passes the check and will not give a warning about this issue:

#include <stdlib.h>

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

void func(size_t 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 (size_t i = 0; i < array_size; ++i) {
        structP->data[i] = 1;
    }
}