Exercise 2.15(b) confusing about cppprimer HOT 6 CLOSED

Do you mean 1.01 (double literal) is not an object,so the reference can't be bound to 1.01?

yes.

But const int &i=1.01 is valid

const reference is different from plain reference.

const int &i = 1.01; // i refers to temporary with value 2.0

Thus, in your example, the const reference is not bound to double literal, rather than the temporary(object).

please check:

http://en.cppreference.com/w/cpp/language/reference_initialization
http://stackoverflow.com/a/2088337/1155235

FYI

from cppprimer.

Hi, @pezy,
I got a question about a reference you have mentioned above. At the end of his answer, he asserted that:

... Any name which contains adjacent underscores is reserved to the C++ implementation.

While I try to compile codes like:

int main()
{
    int __A = 9;
    int const& ref = __A;
    int __B = ref + 6;
    int const& another_ref = __B;
    int test__variable_name = 88;
    std::cout << ref << " " << another_ref << " " << test__variable_name << std::endl;
    return 0;
}

gcc-c++ compiler (version:4.9.2) don't throw a(n) warning or error to me as I expect and its outputs are correct. And of course I know that assign this kind of variable name is inappropriate, I just wanna know whether modern C++ explicitly forbid such behaviors or not.

from cppprimer.

I just wanna know whether modern C++ explicitly forbid such behaviors or not.

No.

It's just a coding style... (established by usage)
You can free to use underscores to start your variable names, but we do not recommend doing this, cause the style is a "privilege" of compiler implementation.

from cppprimer.

@pezy Got it and thank you. I found that almost, well actually every variable name I can find in C++ headers start with underscores, like _Ptr or _Rb_tree_node . Is that what you guys call "compiler implementation" or maybe I just make a mistake?

from cppprimer.

Is that what you guys call "compiler implementation"

Exactly. 😄

from cppprimer.

@pezy A big thanks!

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