Why is C++11 constexpr so restrictive?

Question

As you probably know, C++11 introduces the constexpr keyword.

C++11 introduced the keyword constexpr, which allows the user to guarantee that a function or object constructor is a compile-time constant. [...] This allows the compiler to understand, and verify, that [function name] is a compile-time constant.

My question is why are there such strict restrictions on form of the functions that can be declared. I understand desire to guarantee that function is pure, but consider this:

The use of constexpr on a function imposes some limitations on what that function can do. First, the function must have a non-void return type. Second, the function body cannot declare variables or define new types. Third, the body may only contain declarations, null statements and a single return statement. There must exist argument values such that, after argument substitution, the expression in the return statement produces a constant expression.

That means that this pure function is illegal:

constexpr int maybeInCppC1Y(int a, int b)
{
    if (a>0)
        return a+b;
    else
        return a-b;
  //can be written as   return  (a>0) ? (a+b):(a-b); but that isnt the point
}

Also you cant define local variables... :( So I'm wondering is this a design decision, or do compilers suck when it comes to proving function a is pure?

Many years later, in c++14, I think it is compilable now. (coliru.stacked-crooked.com/a/e84489e166f931b7) or do I misunderstand something? Is it undefined behavior? — cppBeginner
c++14 relaxed rules for constexpr, but this question is about C++11 — NoSenseEtAl
True, the answers and pretext apply to c++11, but even today, this is the first result when searching for "c++ constexpr unroll for". Future searcher, please note that modern c++ does allow a for loop within a constexpr function/constructor. en.cppreference.com/w/cpp/language/constexpr — memtha

Daniel Earwicker Daniel Earwicker · Accepted Answer · 2011-11-29T10:20:45

The reason you'd need to write statements instead of expressions is that you want to take advantage of the additional capabilities of statements, particularly the ability to loop. But to be useful, that would require the ability to declare variables (also banned).

If you combine a facility for looping, with mutable variables, with logical branching (as in if statements) then you have the ability to create infinite loops. It is not possible to determine if such a loop will ever terminate (the halting problem). Thus some sources would cause the compiler to hang.

By using recursive pure functions it is possible to cause infinite recursion, which can be shown to be equivalently powerful to the looping capabilities described above. However, C++ already has that problem at compile time - it occurs with template expansion - and so compilers already have to have a switch for "template stack depth" so they know when to give up.

So the restrictions seem designed to ensure that this problem (of determining if a C++ compilation will ever finish) doesn't get any thornier than it already is.

Why is C++11 constexpr so restrictive?

5 Answers