Why we need to return reference to istream/ostream while overloading >> and

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The reason is a combination of several facts.

  1. 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.

  2. Since you want your operator to work on any istream, i.e. any object derived from std::istream, you cannot define

    operator<<(istream_type, object);    // take istream by value

    since this would only work for the specific istream type istream_type, but not for a generic istream. 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 from std::istream).

  3. 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<< a template and take and return the istream_type by value, but that requires the istream type to have a copy constructor, which it may well not have for good reasons.

  4. 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).

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