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As far as I understand it, constexpr can be seen as a hint to the compiler to check whether given expressions can be evaluated at compile-time and do so if possible.
I know that it also imposes some restriction on the function or initialization declared as constexpr but the final goal is compile-time evaluation, isn't it?
So my question is, why can't we leave that at the compiler? It is obviously capable of checking the pre-conditions, so why doesn't it do for each expression and evaluate at compile-time where possible?
I have two ideas on why this might be the case but I am not yet convinced that they hit the point:
a) It might take too long during compile-time.
b) Since my code can use constexpr functions in locations where normale functions would not be allowed the specifier is also kind of part of the declaration. If the compiler did everything by itself, one could use a function in a C-array definition with one version of the function but with the next version there might be a compiler-error, because the pre-conditions for compile-time evaluation are no more satisfied.
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Yes ([dcl. constexpr], §7.1. 5/2 in the C++11 standard): "constexpr functions and constexpr constructors are implicitly inline (7.1.
A static constexpr variable has to be set at compilation, because its lifetime is the the whole program. Without the static keyword, the compiler isn't bound to set the value at compilation, and could decide to set it later.
Quick A: constexpr guarantees compile-time evaluation is possible if operating on a compile-time value, and that compile-time evaluation will happen if a compile-time result is needed.
const applies for variables, and prevents them from being modified in your code. constexpr tells the compiler that this expression results in a compile time constant value, so it can be used in places like array lengths, assigning to const variables, etc.
constexpr is not a "hint" to the compiler about anything; constexpr is a requirement. It doesn't require that an expression actually be executed at compile time; it requires that it could.
What constexpr does (for functions) is restrict what you're allowed to put into function definition, so that the compiler can easily execute that code at compile time where possible. It's a contract between you the programmer and the compiler. If your function violates the contract, the compiler will error immediately.
Once the contract is established, you are now able to use these constexpr functions in places where the language requires a compile time constant expression. The compiler can then check the elements of a constant expression to see that all function calls in the expression call constexpr functions; if they don't, again a compiler error results.
Your attempt to make this implicit would result in two problems. First, without an explicit contract as defined by the language, how would I know what I can and cannot do in a constexpr function? How do I know what will make a function not constexpr?
And second, without the contract being in the compiler, via a declaration of my intent to make the function constexpr, how would the compiler be able to verify that my function conforms to that contract? It couldn't; it would have to wait until I use it in a constant expression before I find that it isn't actually a proper constexpr function.
Contracts are best stated explicitly and up-front.
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constexpr can be seen as a hint to the compiler to check whether given expressions can be evaluated at compile-time and do so if possible
No, see below
the final goal is compile-time evaluation
No, see below.
so why doesn't it do for each expression and evaluate at compile-time where possible?
Optimizers do things like that, as allowed under the as-if rule.
constexpr is not used to make things faster, it is used to allow usage of the result in context where a runtime-variable expression is illegal.
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This is only my evaluation, but I believe your (b) reason is correct (that it forms part of the interface that the compiler can enforce). The interface requirement serves both for the writer of the code and the client of the code.
The writer may intend something to be usable in a compile-time context, but not actually use it in this way. If the writer violates the rules for constexpr, they might not find out until after publication when clients who try to use it constexpr fail. Or, more realistically, the library might use the code in a constexpr sense in version 1, refactor this usage out in version 2, and break constexpr compatibility in version 3 without realizing it. By checking constexpr-compliance, the breakage in version 3 will be caught before deployment.
The interface for the client is more obvious --- an inline function won't silently become constexpr-required because it happened to work and someone used that way.
I don't believe your (a) reason (that it could take too long for the compiler) is applicable because (1) the compiler has to check much of the constexpr constraints anyway when the code is marked, (2) without the annotation, the compiler would only have to do the checking when used in a constexpr way (so most functions wouldn't have to be checked), and (3) IIUC the D programming language actually does allow functions to be compile-time evaluated if they meet requirements without any declaration assistance, so apparently it can be done.
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