No, using flexible array members in C is not bad practice.

This language feature was first standardized in ISO C99, 6.7.2.1 (16). In the following revision, ISO C11, it is specified in Section 6.7.2.1 (18).

You can use them like this:

struct Header {
    size_t n  ;
    long   v[];
};
typedef struct Header Header;

size_t  n = 123; // can dynamically change during program execution
// ...
Header *h = malloc(sizeof(Header) + sizeof(long[n]));
h->n      = n;

Alternatively, you can allocate like this:

Header *h = malloc(sizeof *h + n * sizeof h->v[0]);

Note that sizeof(Header) includes eventual padding bytes, thus, the following allocation is incorrect and may yield a buffer overflow:

Header *h = malloc(sizeof(size_t) + sizeof(long[n])); // invalid!

A struct with a flexible array members reduces the number of allocations for it by 1/2, i.e. instead of 2 allocations for one struct object you need just 1. Meaning less effort and less memory occupied by memory allocator bookkeeping overhead. Furthermore, you save the storage for one additional pointer. Thus, if you have to allocate a large number of such struct instances you measurably improve the runtime and memory usage of your program (by a constant factor).

In contrast to that, using non-standardized constructs for flexible array members that yield undefined behavior (e.g. as in long v[0]; or long v[1];) obviously is bad practice. Thus, as any undefined-behaviour this should be avoided.

Since ISO C99 was released in 1999, more than 20 years ago, striving for ISO C89 compatibility is a weak argument.

C++ doesn't support C99 flexible array members at the end of structures, either using an empty index notation or a 0 index notation (barring vendor-specific extensions):

struct blah
{
    int count;
    int foo[];  // not valid C++
};

struct blah
{
    int count;
    int foo[0]; // also not valid C++
};

As far as I know, C++0x will not add this, either.

However, if you size the array to 1 element:

struct blah
{
    int count;
    int foo[1];
};

the code will compile, and work quite well, but it is technically undefined behavior. You can allocate the appropriate memory with an expression that is unlikely to have off-by-one errors:

struct blah* p = (struct blah*) malloc( offsetof(struct blah, foo[desired_number_of_elements]);
if (p) {
    p->count = desired_number_of_elements;

    // initialize your p->foo[] array however appropriate - it has `count`
    // elements (indexable from 0 to count-1)
}

So it's portable between C90, C99 and C++ and works just as well as C99's flexible array members.

Raymond Chen did a nice writeup about this: Why do some structures end with an array of size 1?

Note: In Raymond Chen's article, there's a typo/bug in an example initializing the 'flexible' array. It should read:

for (DWORD Index = 0; Index < NumberOfGroups; Index++) { // note: used '<' , not '='
  TokenGroups->Groups[Index] = ...;
}