CODEWITHSUNDEEP
cconstantsvariable-length-array
Raman
Raman

Reputation: 2785

Array size of enum length gives compilation error

Have a look at the code below,

This code compiles fine:

enum ids {
    X,
    Y,
    NUM_IDS,
}

void some_func(void)
{
    static char* const names[NUM_IDS] = { "name X" , "name Y"};
}

However this code doesn't compile: error: storage size of 'names' isn't constant

enum ids {
    X,
    Y,
    NUM_IDS,
}

void some_func(void)
{
    int nr_names = NUM_IDS;
    static char* const names[nr_names] = { "name X" , "name Y"};
}

I think I misunderstood the meaning of a constant expression. Is it the case that in second way it is becoming a VLA which is not present in C90? Somebody please clarify.

Upvotes: 1

Views: 109

Answers (2)

alinsoar
alinsoar

Reputation: 15793

The problem is that VLAs are not allowed to have static storage duration

Quoting from ISO 9899 6.7.5.2 Array declarators

If an identifier is declared to be an object with static storage duration, it shall not have a variable length array type.

So, you are not allowed to declare a static VLA in general. Because nr_names is not constant, you are not allowed to use static in your second code.

The first code is correct, as NUM_IDS is constant expression.

Other problem with your second code is that a VLA cannot be initialized.

I quote from ISO 9899 6.7.8 Initialization

The type of the entity to be initialized shall be an array of unknown size or an object type that is not a variable length array type

Upvotes: 1

Blaze
Blaze

Reputation: 16876

static char* const names[nr_names] is a VLA because nr_names isn't a constant expression, but a (non-const) int. Sure, in this short example it's always equal to NUM_IDS, but you still can't do that.

On an unrelated side note, it is recommended that the char is defined as const, as modifying it won't work because it's part of the program's binary (in C++ it wouldn't let you have it non-const):

static const char* const names[NUM_IDS] = { "name X" , "name Y" };

Upvotes: 3

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