As far as I know, the const
qualifier in C++ basically declares internal linkage, and sometimes it allows the variable to be used as a constant expression so that it can be placed into array bounds, switch cases, etc.
But apparently this is not always the case, and I have no clue regarding the proper usage of const
and constexpr
.
Specifically, I found that when an element of a const-qualified array is used in array bounds it is not treated as a constant expression, as seen in the following code.
const int N = 3; int foo[N] = {1, 2, 3}; // Valid const int bar[5] = {1, 2, 3, 4, 5}; int arr[bar[2]] = {1, 2, 3}; // Invalid because a VLA can't be initialized
Using constexpr
instead of const
in the later part solves the problem. But why is the last statement invalid? And exactly what is required for an expression to be constant?
We use the const qualifier to declare a variable as constant. That means that we cannot change the value once the variable has been initialized. Using const has a very big benefit. For example, if you have a constant value of the value of PI, you wouldn't like any part of the program to modify that value.
The const keyword can be used as a qualifier when declaring objects, types, or member functions. When qualifying an object, using const means that the object cannot be the target of an assignment, and you cannot call any of its non-const member functions.
The const keyword specifies that a variable's value is constant and tells the compiler to prevent the programmer from modifying it.
A constant expression is an expression that contains only constants. A constant expression can be evaluated during compilation rather than at run time, and can be used in any place that a constant can occur.
The only time const
means the same thing as constexpr
in the declaration of a variable, is when the variable is of integral or enumeration type. In addition, the initializer of this variable declaration must be a constant expression. e.g.
const int n = 42; // same as constexpr // type is int // initializer is integer literal, which is constant expression std::cin >> x; // some user input const int n = x; // NOT constexpr // because initializer is not a constant expression const double n = 4.2; // NOT constexpr // type is not integral or enumeration type
Your last line of code fails to compile because bar
is not an integral or enumeration type, and hence it's not a constexpr
. Since it's not constexpr
none of its elements are constexpr
either, and hence they cannot be used as an array bound.
The reason for this special case with integers is historical: array bounds need to be constant expressions, but before C++11, the only way to express that was with a const int
. Technically, the rules could be changed to require the declaration to have constexpr
but that would break exisiting code and so it won't be changed.
And exactly what is required for an expression to be constant?
This is interesting, because the language doesn't actually say what is required for an expression to be a constant expression. Instead, it assumes that all expressions are constant expressions, and provides a list of conditions that if not satisfied will make the expression not a constant expression.
The rule is here:
An expression E is a core constant expression unless the evaluation of E, following the rules of the abstract machine ([intro.execution]), would evaluate one of the following: ...
and this is followed by a list of conditions that make an expression not a constant expression.
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