Why we need to return reference to istream/ostream while overloading >> and << operators?
What happens if I do not return din
or开发者_JAVA技巧 dout
, actually I'm reading a book in which writer returns back stream references
istream & operator>>(istream &din,vector &a)
{
for(int i=0;i<size;i++)
din>>a.v[i];
return din;
}
ostream & operator<<(ostream &dout,vector &a)
{
dout<<"("<<a.v[0];
for(int i=1;i<size;i++)
dout<<", "<<a.v[i];
dout<<")";
return dout;
}
The reason is a combination of several facts.
You want to be able to chain input and output operations as in
in >> x >> y; out << z << std::precision(10) << t << std::endl;
so you must return something that allows
operator<<
again.Since you want your operator to work on any
istream
, i.e. any object derived fromstd::istream
, you cannot defineoperator<<(istream_type, object); // take istream by value
since this would only work for the specific istream type
istream_type
, but not for a genericistream
. For that one must use polymorphism, i.e. either take a reference or a pointer (which will be a reference or pointer to a class derived fromstd::istream
).Since you only have a reference to the istream, you cannot return the istream object itself (which may be of a type not even defined at the point of the definition of
operator<<
) but only the reference you've got.One could get around this restriction by defining
operator<<
atemplate
and take and return theistream_type
by value, but that requires theistream
type to have a copy constructor, which it may well not have for good reasons.In order to envoke polymorphism one could, in principle, use pointers (to streams) rather than references. However,
operator<<(stream*,const char*)
is not allowed in C++ (at least one operand must be of class or enumeration type).Thus, with stream pointers one must use function-call syntax and you're back with C-style
fprintf(stream*, args...)
.Moreover, pointers can be null or dangling, which in fact is their default state (when declared without initializer), while a reference can be assumed to be valid (it cannot be declared without initializer).
In this case when the reference is returned you can combain the operator in a chain. For example
std::cout << "Hello " << "Rajat Verma";
This is equivalent to the following calls of the operator
operator <<( operator <<( std::cout, "Hello" ), "Rajat Verma" );
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
returns reference to std::cout
One more thing is that ostream and istream standard objects such as cout and cin use a private copy constructors so they should be returned by reference not by value
When you type: cout << vector; cout has the type of ostream so when you use " << " it does need to return an arguement with ostream type for cout to work
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