CODEWITHSUNDEEP
c++
AlwaysLearning
AlwaysLearning

Reputation: 8001

const (but not constexpr) used as the built-in array size

I understand that the size of the built-in array must be a constant expression:

// Code 1
constexpr int n = 5;
double arr[n];

I do not understand why the following compiles:

// Code 2
const int n = 5;
double arr[n]; // n is not a constant expression type!

Furthermore, if the compiler is smart enough to see that n is initialized with 5, then why does the following not compile:

// Code 3
int n = 5;
double arr[n]; // n is initialized with 5, so how is this different from Code 2?

P.S. This post answers using quotes from the standard, which I do not understand. I will very much appreciate an answer that uses a simpler language.

Upvotes: 3

Views: 100

Answers (3)

eerorika
eerorika

Reputation: 238281

n is not a constant expression type!

There is no such thing as a constant expression type. n in that example is a expression, and it is in fact a constant expression. And that is why it can be used as the array size.

It is not necessary for a variable to be declared constexpr in order for its name to be a constant expression. What constexpr does for a variable, is the enforcement of compile time constness. Examples:

int a = 42;

Even though 42 is a consant expression, a is not; Its value may change at runtime.

const int b = 42;

b is a constant expression. Its value is known at compile time

const int c = rand();

rand() is not a constant expression, and so c is neither. Its value is determined at runtime, but may not change after initialisation. 

constexpr int d = 42;

d is a constant expression, just like b.

constexpr int f = rand();

Does not compile, because constexpr variables must be initialised with a constant expression.

then why does the following not compile:

Because the rules of the language don't allow it. The value of n is not compile time constant. The value of a non-const variable can change at runtime.

The language cannot have a rule that some value doesn't change at runtime, then it is a constant expression. That would not be of any use to the programmer since they cannot assume which compiler will be able to prove the constness of which variable.

The language has to exactly specify the cases where an expression is constant. It would also be infeasible to specify that a non-const variable is a constant expression if it hasn't been modified before its use, because it is impossible to prove in most cases, even though you've found one case where the proof happens to be easy.

Upvotes: 4

Theodor Badea
Theodor Badea

Reputation: 464

A value declared constexpr means that this value does not change and is known during compile time.

A value declared const means that this value does not change after initialization, but it is not mandatory to be known during compile time. In other words, a constexpr is const, but a const is not constexpr.

Your "Code 3" example doesn't work because you need a constant known at compile time in order to allocate memory for a vector, so you need a constexpr.

Upvotes: 0

NathanOliver
NathanOliver

Reputation: 180415

// n is not a constant expression type!

But it is. Per [expr.const]/3

A variable is usable in constant expressions after its initializing declaration is encountered if it is a constexpr variable, or it is a constant-initialized variable of reference type or of const-qualified integral or enumeration type. An object or reference is usable in constant expressions if it is [...]

  • a complete temporary object of non-volatile const-qualified integral or enumeration type that is initialized with a constant expression.

So, if you have a const integer intialized with a constant expression then you still have a constant expression as nothing can change. This is a rule that existed before constexpr was ever a thing as it allowed programmers to initialize arrays with constant variables instead of using macros.

Furthermore, if the compiler is smart enough to see that n is initialized with 5, then why does the following not compile:

Because the integer is not const so it could be changed. Even though in your case you can prove it can't change, in general you can't so it is just not allowed.

Upvotes: 1

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