Testing stream.good() or !stream.eof() reads last

问题:

Possible Duplicate:
Why is iostream::eof inside a loop condition considered wrong?

I have the following piece of code:

ifstream f(\"x.txt\");
string line;
while (f.good()) {
  getline(f, line);
  // Use line here.
}

But this reads the last line twice. Why does this happen and how do I fix it?

Something very similar happens with:

ifstream f(\"x.txt\");
string line;
while (!f.eof()) {
  getline(f, line);
  // Use line here.
}

回答1:

You very, very rarely want to check bad, eof, and good. In particular for eof (as !stream.eof() is a common mistake), the stream currently being at EOF does not necessarily mean the last input operation failed; conversely, not being at EOF does not mean the last input was successful.

All of the stream state functions – fail, bad, eof, and good – tell you the current state of the stream rather than predicting the success of a future operation. Check the stream itself (which is equivalent to an inverted fail check) after the desired operation:

if (getline(stream, line)) {
  use(line);
}
else {
  handle_error();
}

if (stream >> foo >> bar) {
  use(foo, bar);
}
else {
  handle_error();
}

if (!(stream >> foo)) {  // operator! is overloaded for streams
  throw SomeException();
}
use(foo);

To read and process all lines:

for (std::string line; getline(stream, line);) {
  process(line);
}

Pointedly, good() is misnamed and is not equivalent to testing the stream itself (which the above examples do).

回答2:

Just use

ifstream f(\"x.txt\");
while (getline(f, line)) {
    // whatever
}

This is the idiomatic way to write such a loop. I\'ve not been able to reproduce the error (on a Linux machine).

回答3:

It didn\'t read the last line twice but because it failed to read when it reached eof, your string line has the value it had previously.

That is because f is no longer \"good\" when it has read EOF, not when it is about to read it.