What is the purpose of marking the set function (s

问题:

This question already has answers here:

What is the use of a constexpr on a non-const member function? (2 answers)

Closed last year.

I cannot understand the purpose of marking the setter function as constexpr, that is allowed since C++14. My misunderstanding comes from the next situation: I declare a class with a constexpr c-tor, and I am about to use it in a constexpr context, by creating a constexpr instance of that class constexpr Point p1. An object p1 now is constant and its value could not be changed, so the constexpr setter could not be called. On the other hand, when I create an instance of my class Point in a non-constexpr context Point p, I can call the setter for that object, but now setter will not execute at compile-time, because the object is not constexpr!

As a result, I do not understand how can I enhance the performance of my code using constexpr for setters.

This is the code that demonstrates calling a constexpr setter on an non-constexpr object, that means run-time computation, instead of the compile-time:

class Point {
public:
    constexpr Point(int a, int b)
    : x(a), y(b) {}

    constexpr int getX() const noexcept { return x; }
    constexpr int getY() const noexcept { return y; }

    constexpr void setX(int newX) noexcept { x = newX; }
    constexpr void setY(int newY) noexcept { y = newY; }
private:
    int x;
    int y;
};


int main() {
    Point p{4, 2};
    constexpr Point p1{4, 2};

    p.setX(2);
}

Could anyone help me to understand what is the purpose of marking the setter function as constexpr?

回答1:

Basically it is nice when you have to deal with constexpr function.

struct Object {
  constexpr void set(int n);
  int m_n = 0;
};

constexpr Object function() {
   Object a;
   a.set(5);
   return a;
}

constexpr Object a = function();

The idea is to be able to perform compile time initialization within another functions that will be executed at the compile time. It is not done to be applied on constexpr object.

Another things to know is that constexpr member functions are not const member functions since C++14 :).

回答2:

The need arise with new constexpr rule with C++14: inside constexpr function, you can now use multiple statements, including for loops and control flow.

Here's an example:

constexpr int count5(int start) {
    int acc = 0;

    for (int i = start ; i<start+5 ; ++i) {
        acc += i;
    }

    return acc;
}

constexpr int value = count5(10); // value is 60!

As you can see, we can do many mutation to variable in a constexpr context. The compiler becomes like an interpreter, and as long as the result of the constexpr fonction is consistent and you don't mutate already computed constexpr variables, it may mutate the values during the interpretation.

回答3:

A function with constexpr qualifier will evaluate the return of the function at compile time, which can significantly boost performance of a program (no extra computation, no instruction counter jumping, etc.). There are a few requirements to qualify a function thusly, so check out this explanation from IBM.