C++ Primer exercise 2.27 [5th ed.]

Question

I am doing exercise 2.27 from C++ primer 5th edition and I am confused in this question:

Exercise: Which of the following initializations are legal? Explain why.

(c) const int i = -1, &r = 0;

I came to conclusion that r is illegal because this will be same as below:

const int i = -1;
int &r = 0;

But this github repo suggest that (c) is same as below:

const int i = -1;
const int &r = 0;

So, it contradicts to my answer, please provide me the correct answer.

P.S.: I am begineer in C++ language.

_" I am begineer in C++ language."_ Then Stack Overflow is probably the wrong place for you. — πάντα ῥεῖ, Aug 06 '16 at 13:12
What makes you think that answer might be wrong? This question comes across as simply you choosing to disregard that answer, because accepting it would mean you were wrong. — , Aug 06 '16 at 13:12
You can check whether the compiler agrees with you by using `std::is_same` and `decltype`: http://coliru.stacked-crooked.com/a/c0bb0ba07726c1e6 — milleniumbug, Aug 06 '16 at 13:13
@πάνταῥεῖ Beginners can certainly ask beginner questions here. It's just that they have to be *good* beginner questions, by the same standards of *good* as get applied to the rest of us. — , Aug 06 '16 at 13:13
Just like the type `int`, the qualifier `const` is for all variabled defined in the same declaration statement. — Some programmer dude, Aug 06 '16 at 13:13
@hvd You're right, but such cases are rare from my experience. — πάντα ῥεῖ, Aug 06 '16 at 13:14
A tip for beginners (and everyone else) is to avoid declarations like `const int i = -1, &r = 0;`. Just declare each variable on a separate line, and you don't have to wonder what it means. — Bo Persson, Aug 06 '16 at 13:21

score 1 · Accepted Answer · answered Aug 06 '16 at 13:55

1

The type specifier (int) with the qualifier (const) belong to all declarators in the declaration

const int i = -1, &r = 0;

Thus declarators i and &r have the type specifier and qualifier const int. Moreover you may not write for example

int &r = 0;

because a temporary object (in this case expression 0) may not be bound to a non-constant reference.

However you could write

int &&r = 0;

answered Aug 06 '16 at 13:55

Vlad from Moscow

does this also apply to `volatile` qualifier? – silverwolf Aug 06 '16 at 14:05
@silverwolf This is applied to cv qualifiers. – Vlad from Moscow Aug 06 '16 at 14:06

1 Answers1