I'm developing a C library, there are two functions: func1 and func2.

I want to generate an compile-time error if developer forgets to call func1 before calling func2:

This will be OK:

func1();
func2();  // OK

But this would fail:

func2();  // ERROR, you forget to call func1()

Of course, it's very easy to generate a runtime error but I would like to generate a compile-time error.

I've tried as below but it's not working because I can't modify a constexpr variable:

static constexpr bool b {false};

void func1() {
    b = true; // ERROR!
}

typename<std::enable_if_t<b == true>* = nullptr>
void func2() {}

I'm not very good at metaprogramming. I want to know if it's possible to generate a compile-time error in this case.

CodePudding user response：

It is not possible at least in corner cases. Look at the following code (headers and test for scanf return value omitted for brievety):

int main() {
    int code;
    printf("Enter code value: ");
    scanf("%d", &code);
    if (code > 0) func1();
    func2();               // has func1 be called ?
    return 0;
}

Here, even the best static analysis tool will be unable to state whether func1 is called before func2 because it depends on a value which will only be known at runtime. Ok, this is a rather stupid example, but real world programs often read their config data at run time and that config can change even part of their initialization.

CodePudding user response：

Make temporal coupling explicit:

• make func1 return a type which should be consumed by func2:

struct func1_result {
// Possibly constructors private, and friendship to `func1`.
//...
};

func1_result func1();
void func2(const func1_result&);

So you cannot call func2 without creating first func1_result (returned by func1).