How can std::cin return a bool and itself at the same time?

Question

I'm reading a book on C++ that says that if I use the >> operator it returns the object at the left side of the operator so in this example

std::cin >> value1; 

the code returns std::cin.

But if I do this

while(std::cin >> value1) 

My code will be in the loop until there is a std::cin error so that must mean that the operator returns a bool that is true when std::cin does not fail and false when std::cin fails.

Which is one is it?

Answer 1

[...] so that must mean that the operator returns a bool [...]

Nope, it returns std::cin (by reference). The reason why while(std::cin >> value); works is because std::istream (which is the type of std::cin) has a conversion operator.

A conversion operator basically permits a class to be implicitly (if it is not marked explicit) to be converted to a given type. In this case, std::istream defines its operator bool to return whether an error (typically failbit) occurred;

_{[std::ios_base::operator bool()]}

Returns true if the stream has no errors and is ready for I/O operations. Specifically, returns !fail().

---

explicit operator bool() const; 

Note than even though the operator is explicit (which shouldn't allow implicit conversions like if (std::cin);), the qualifier is ignored when used in a context that requires a bool, like if, while loops and for loops. Those are exceptions though, not rules.

Here is an example:

if (std::cin >> value);  //OK, a 'std::istream' can be converted into a 'bool', which                           //therefore happens implicitly, without the need to cast it:   if (static_cast<bool>(std::cin >> value)); //Unnecessary  bool b = std::cin >> value;  //Error!! 'operator bool' is marked explicit (see above), so                               //we have to call it explicitly:  bool b = static_cast<bool>(std::cin >> value); //OK, 'operator bool' is called explicitly