Programming principles and practice using C++ error: constexpr

Question

In Stroustrup's "Programming principles and practice" book, there's an example of constexpr like this:

void user(Point p1)
{
    Point p2 {10,10};
    Point p3 = scale(p1); // OK: p3 == {100,8}; run-time evaluation is fine
    constexpr Point p4 = scale(p2); // p4 == {100,8}
    constexpr Point p5 = scale(p1); // error: scale (p1) is not a constant
                                    // expression
    constexpr Point p6 = scale(p2); // p6 == {100,8}
    // . . .
}

But think he is mistaken: p2although initialized with constant expression arguments (literals here 10, 10) it is not a constexpr object because it is not declared so.

So normally p4 and p6 are in error here: (cannot use p2 in a constant expression). it is like p1.

To correct it:
```
 constexpr Point p2{10, 10};
```

Note that Stroustrup's books have plenty of typos. The errata only mention those he doesn't consider obvious. — molbdnilo
– molbdnilo, Commented Jul 29, 2022 at 10:27

Goswin von Brederlow · Accepted Answer · 2022-07-29 16:26:26Z

1

You know who is really good at telling you if something is allowed as a constexpr? Your compiler. https://godbolt.org/z/4Kdocx83v

And you are right, it's broken.

answered Jul 29, 2022 at 16:26

Goswin von Brederlow

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2 Comments

Ben Voigt Over a year ago

compiler constexpr support isn't stable (they're chasing a moving target), so compiler bugs are not uncommon. Therefore while it's useful to know what a particular compiler thinks of a particular piece of code, you shouldn't accept it as absolute truth.

Goswin von Brederlow Over a year ago

@BenVoigt Nothing is ever absolute. But I doubt there is still much happening with constexpr and c++11. The moving target is in support for c++20 and the STL.

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