Question

const int i = 100;
int *j = &i;
int array[i] = {0};

Is this legal? I read somewhere that without the &i, i and 100 would be added to the symbol-table, but because of the &i, storage is forced for i, and 100 would be stored in i at compile time; therefore the compiler would not be able to read the value of i (from storage) to allocate the array – is this true?

Answer 1

Is this legal? I read somewhere that without the &i, i and 100 would be added to the symbol-table, but because of the &i, storage is forced for i, and 100 would be stored in i at compile time; therefore the compiler would not be able to read the value of i (from storage) to allocate the array – is this true?

Here’s the language from the C99 standard:

6.7.3 Type qualifiers
…
3 The properties associated with qualified types are meaningful only for expressions that
are lvalues.¹¹⁴⁾

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^{114) The implementation may place a const object that is not volatile in a read-only region of
storage. Moreover, the implementation need not allocate storage for such an object if its address is
never used.}

So, sort of true, but not really relevant in this case. The compiler doesn’t have to create storage for i to use its value elsewhere; it may simply use an immediate operand with the value of 100 in place of i. Either way, the line int j = &i; is unnecessary.

i is not an integer constant expression, meaning it’s not a compile-time constant (in C; C++ is different in this regard); it’s a run-time variable whose value may not be modified during its lifetime.

As of C99, you can specify array sizes using a run-time variable, even one declared const, so

const int i = 100;
int array[i];

will allocate array as a 100-element array of int. However, you cannot use an initializer with a VLA, so int array[i] = {0}; is not valid.

Again, C++ is different, and doesn’t support VLAs at all. But since i is declared const, C++ treats it as a compile-time constant, meaning your code should be legal C++ (it builds for me, anyway).

More from C99:

6.7.5.2 Array declarators
…
4 If the size is not present, the array type is an incomplete type. If the size is * instead of
being an expression, the array type is a variable length array type of unspecified size,
which can only be used in declarations with function prototype scope;¹²⁴⁾ such arrays are
nonetheless complete types. If the size is an integer constant expression and the element type has a known constant size, the array type is not a variable length array type;
otherwise, the array type is a variable length array type.
…
6.7.8 Initialization
…
3 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.

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^{124) Thus, * can be used only in function declarations that are not definitions (see 6.7.5.3).}