Thread: If EOF evaluates to -1 why does it break a while loop

Originally Posted by Vespasian_2

In the conditional statement in the while loop below, the extraction operator is used on an input stream object "i". When the file reaches an EOF, the loop breaks.

This doesnt make sense as I understand it EOF usually evaluates to -1 and a loop only breaks on a value of false or 0.

Code:

while (i >> next){
//do stuff...
}

How can you explain this?

I suggest asking C++ questions in the C++ sub-forum; you are more likely to get faster answers.
Edit: In case you did not know the code posted is C++ code!

Tim S.

EOF is just a macro defined to be a negative integer, usually -1. It happens to be returned by some functions to indicate the end of file condition, but there is nothing special about it.

Assuming i is an istream, the extraction operator returns a reference to the istream object itself so that extraction operators can be chained. E.g., due to the left-to-right associativity of the >> operator, i >> first >> next >> last is processed like (((i >> first) >> next) >> last), with (i >> first) returning a reference to i so that the next operation is (i >> next), etc.

Finally, when the final istream reference is returned and evaluated in a boolean context, it returns false if it's at eof or in an error condition.

Code:

#include <iostream>
using namespace std;
 
class Object {
    int n;
public:
    Object(int n) : n{n} {}
 
    operator bool() { return n % 2 == 0; } // true if n is even
 
    friend Object& chain(Object& o, int n) {
        cout << '+' << n << '\n';
        o.n += n;
        return o;
    }
     
    friend ostream& operator<<(ostream& os, const Object& o) {
        return os << o.n;
    }
};
 
int main() {
    Object o(11), p(22);
 
    if (o) cout << "o\n";
    if (p) cout << "p\n";
 
    cout << o << '\n';
    chain(chain(chain(o, 1), 2), 3);
    cout << o << '\n';
}

If you do want to write your own conversion function to bool for a similiar purpose, it would be better to declare it as explicit to avoid unintended implicit conversions. It typically would also be declared const as it is unlikely that the conversion will change the observable state of the object:

Code:

explicit operator bool() const { /* ... */ }