CODEWITHSUNDEEP
cvariable-length-array
Jack
Jack

Reputation: 16724

ISO C90 forbids variable length array

I'm dynamically calculating the size of an array. Something like:

void foo(size_t limit)
{
  char buffer[limit * 14 + 1];
}

But just GCC compiler says:

error: ISO C90 forbids variable length array ‘buffer’

searching on SO I found this answer:

C99 §6.7.5.2:

If the size is an expression that is not an integer constant expression... ...each time it is evaluated it shall have a value greater than zero.

So, I did the re-declaration of size limit type variable to:

void foo(const size_t limit)

But it continues to give warning for me. Is this a GCC bug?

Upvotes: 6

Views: 37891

Answers (6)

ouah
ouah

Reputation: 145839

const does not introduce a constant in C but a read-only variable.

#define SIZE 16
char bla[SIZE];   // not a variable length array, SIZE is a constant

but

const int size = 16;
char bla[size];   // C99 variable length array, size is a constant

Upvotes: 6

P.P
P.P

Reputation: 121397

C90 doesn't allow variable length arrays. However, you can use the c99-gcc compiler to make this work.

You are compiling with c90-gcc but looking at C99 specifications.

Upvotes: 4

Christoph
Christoph

Reputation: 169603

const-qualifying a variable doesn't make it a compile-time constant (see C99 6.6 §6 for the defintion of an integer constant expression), and before the introduction of variable-length arrays with C99, array sizes needed to be compile-time constants.

It's rather obvious that const-qualify a variable doesn't make it a compile-time constant, in particular in case of function parameters which won't be initialized until the function is called.

I see the following solutions to your problem:

  • compile your code as C99 via -std=c99 or -std=gnu99
  • allocate your buffer via malloc()
  • use alloca() if available, which is the closest you can come to variable-length arrays with C90
  • choose a maximum buffer size which is always used and fail if the given limit argument overflows

As a side note, even though C99 allows variable-length arrays, it's still illegal to use the value of an integer variable with static storage duration as size for an array with static storage duration, regardless of const-qualification: While there's nothing which prevents this in principle if the integer variable is initialized in the same translation unit, you'd have to special-case variables with visible defintion from those whose definition resides in a different translation unit and would either have to disallow tentative defintions or require multiple compilation passes as the initialization value of a tentatively defined variable isn't known until the whole translation unit has been parsed.

Upvotes: 14

Jens Gustedt
Jens Gustedt

Reputation: 78923

A const qualified variable is not an integer constant expression in the sense of the standard. This has to be a literal constant, an enumeration constant, sizeof or some expression composed with these.

Switch to C99 if you may. The gcc option is -std=c99 (or gnu99 if you want gnu extension.)

Upvotes: 0

MByD
MByD

Reputation: 137342

As written in your question, this is from C99, not C90, you need to compile it against C99 to be able to use variable length arrays.

Upvotes: 0

David Heffernan
David Heffernan

Reputation: 612993

No it is not a bug. You can't use a VLA in C90. When you declared

const size_t limit

that is not a constant expression. A constant expression would be something like a literal value 666.

Note that C differs significantly from C++ in this regard. Even a constant like this

const int i = 666;

is not a constant expression in C. This is the primary reason why constant values are typically declared with #define in C.

Upvotes: 2

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