I\'m dynamically calculating the size of an array. Something like:
void foo(size_t limit)
{
char buffer[limit * 14 + 1];
}
But just GCC c
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.
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 not constant
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:
-std=c99
or -std=gnu99
malloc()
alloca()
if available, which is the closest you can come to variable-length arrays with C90limit
argument overflowsAs 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.
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.
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.
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.)