Re: Why extracting string from stringstream(a) fails?
Maxim Yegorushkin wrote:
Victor Bazarov wrote:
Maxim Yegorushkin wrote:
Alf P. Steinbach wrote:
* Maxim Yegorushkin:
On 25/10/09 18:13, Johannes Schaub (litb) wrote:
Maxim Yegorushkin wrote:
Here is another trick to turn a temporary into an l-value, so that
any
operator>> can work on a temporary stream object:
template<class T>
inline T& lvalue(T const& t) {
return const_cast<T&>(t);
}
Although this will be ill-formed because it requires the stream to
have a
copy constructor.
Interesting.
My understanding is that a copy constructor is only required when
copy initialization is involved or when a conversion is made to
initialize a function argument. In this case there is no copy
initialization or conversion happening. Therefore, the code must be
well formed.
Could you elaborate you point please?
It's different in C++98 and C++0x (C++03 is just C++98 with
corrections).
The argument passing is defined as copy initialization. And in C++98
the implementation is allowed to make any number of copies of an
rvalue actual argument passed to 'T const&' formal argument, or for
any copy initialization. Which means that the type must provide a
suitable copy constructor. For example, that means that you can't do
this thing with a std::auto_ptr. Or a stream.
Even more interesting.
Given the following declaration:
void foo(int& ref);
Could you explain how ref argument can possibly be copy-initialized
please?
I am guessing the same way 'r2' is a "copy" of 'r1' here:
int a = 42;
int &r1 = a;
int &r2 = r1;
:-)
Very well, this is what I wanted to hear. ;)
In this case no copy constructor is required to copy-initialize a
reference to non-const. By induction, the same should hold true for
references to const
(although they can be initialized with r-values, but only if necessary).
Hence, the above lvalue function template is well-formed.
Comeau online compiles lvalue(std::istringstream(...)) just fine.
Yes, but if you turn off the C++0x mode, it will error out.
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