c++ - istream::peek 奇怪的行为。结束符

Question

我刚刚在 C++ 中遇到了一个奇怪的情况。我在做类似的事情:

istream in;
// ...
in.get();      // get a char (which turns out to be the last)
               // curiously ios::eof bit doesn't get set just yet

c = in.peek(); // attempt to peek, return EOF and now set ios::eof bit

if(c == EOF) {
    // realize shouldn't have gotten the last char for some reason
    in.unget(): // attempt to unget, *fails* (because ios:eof bit was set)
}

现在我很好奇为什么 peek 会设置 eof 位；我觉得这非常不直观。它应该只是偷看而不实际消耗任何东西并且不应该改变流状态。另外，为什么 unget 随后不起作用？当 good() 为 false 或其他情况时，标准是否要求所有操作都为 nop？

Answer 1

in.get();      // get a char (which turns out to be the last)
               // curiously ios::eof bit doesn't get set just yet

这不是“好奇”。当读取由于达到eof而失败时，流的EOF位被设置；它不意味着“最后一次阅读把我们带到了eof”。

c = in.peek(); // attempt to peek, return EOF and now set ios::eof bit

就像现在一样。

Also, why unget subsequently doesn't work? Does the standard mandates all operations to be nop when good() is false or something?

...这就是正在发生的事情。您未能以其他方式定义“不起作用”。

如果您想取消您在第 3 行检索到的那个字符，您必须在到达 EOF 时自己清除流的状态。

istream in;
// ...
in.get();      // assume this succeeds (*)

c = in.peek(); // assume this fails and sets EOF bit

if (!in) {
   in.clear(); // clear stream error state for use
   in.unget(); // "put back" that character (*)
}
else {
   // next char from .get() will be equal to `c`
}