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Enforcing function contract at compile time when possible

Tags:

c++

(this question was inspired by How can I generate a compilation error to prevent certain VALUE (not type) to go into the function?)

Let's say, we have a single-argument foo, semantically defined as

int foo(int arg) {
    int* parg;
    if (arg != 5) {
        parg = &arg;
    }

    return *parg;
}

The whole code above is used to illustrate a simple idea - function returns it's own argument unless the argument is equal to 5, in which case behavior is undefined.

Now, the challenge - modify the function in such a way, that if it's argument is known at compile time, a compiler diagnostic (warning or error) should be generated, and if not, behavior remains undefined in runtime. Solution could be compiler-dependent, as long as it is available in either one of the big 4 compilers.

Here are some potential routes which do not solve the problem:

  • Making function a template which takes it's argument as a template parameter - this doesn't solve the problem because it makes function ineligible for run-time arguments
  • Making function a constexpr - this doesn't solve the problem, because even when compilers see undefined behavior, they do not produce diagnostics in my tests - instead, gcc inserts ud2 instruction, which is not what I want.
like image 664
SergeyA Avatar asked Dec 26 '18 14:12

SergeyA


2 Answers

I got error with constexpr when used in constant expression for:

constexpr int foo(int arg) {
    int* parg = nullptr;
    if (arg != 5) {
        parg = &arg;
    }
    return *parg;
}

Demo

We cannot know that argument value is known at compile type, but we can use type representing value with std::integral_constant

// alias to shorten name. 
template <int N>
using int_c = std::integral_constant<int, N>;

Possibly with UDL with operator "" _c to have 5_c, 42_c.

and then, add overload with that:

template <int N>
constexpr auto foo(int_c<N>) {
    return int_c<foo(N)>{};
}

So:

foo(int_c<42>{}); // OK
foo(int_c<5>{}); // Fail to compile

// and with previous constexpr:
foo(5); // Runtime error, No compile time diagnostic
constexpr auto r = foo(5); // Fail to compile

As I said, arguments are not known to be constant inside the function, and is_constexpr seems not possible in standard to allow dispatch, but some compiler provide built-in for that (__builtin_constant_p), so with MACRO, we can do the dispatch:

#define FOO(X) [&](){ \
    if constexpr (__builtin_constant_p(X)) {\
        return foo(int_c<__builtin_constant_p (X) ? X : 0>{});\
    } else {\
        return foo(X); \
    } \
}()

Demo

Note: Cannot use foo(int_c<X>{}) directly, even in if constexpr, as there is still some syntax check.

like image 164
Jarod42 Avatar answered Oct 21 '22 15:10

Jarod42


gcc/clang/intel compilers support __builtin_constant_p, so you can use something like that:

template <int D>
int foo_ub(int arg) {
    static_assert(D != 5, "error");
    int* parg = nullptr;
    if (arg != 5) {
        parg = &arg;
    }

    return *parg;
}

#define foo(e) foo_ub< __builtin_constant_p(e) ? e : 0 >(e)

these statements produce compile time error:

  • foo(5)
  • foo(2+3)
  • constexpr int i = 5; foo(i);

while all others - runtime segfault (or ub if no nullptr is used)

like image 2
Iłya Bursov Avatar answered Oct 21 '22 16:10

Iłya Bursov