为什么即使流中有一些char，in_avail（）也输出零？ [英] Why does in_avail() output zero even if the stream has some char?

问题描述

#include <iostream>
int main( )
{
   using namespace std;
   cout << cin.rdbuf()->in_avail() << endl;
   cin.putback(1);
   cin.putback(1);
   cout << cin.rdbuf()->in_avail() << endl;
   return 0;
} //compile by g++-4.8.1

我认为这将输出0并2

I think this will output 0 and 2

但是当我运行代码时，它输出0和0，为什么？

but when I run the code, it output 0 and 0, why?

cin.putback（1）;诠释cin >> a;输入12 12;

or if I change cin.putback(1); to int a; cin >> a; with input 12 12;

它仍然输出0和0

推荐答案

必须发生的是，您的 putback 在 streambuf 获取与<$相关的区域中未找到任何房间c $ c> std :: cin （否则将有一个读取位置，而 egptr（）-gptr（）将不可用）零），并且必须通过 pbackfail 进入基础层。

What must have happened is that your putback didn't find any room in the streambuf get area associated with std::cin (otherwise a read position would have been available and egptr() - gptr() would have been non-zero) and must have gone to an underlying layer thanks to pbackfail.

in_avail（ ）将调用 showmanyc（），并且返回零（这是该虚拟函数的默认实现）是安全的，因为这意味着读取可能会阻塞，并且可能会失败，但不能保证两者都执行。显然，在这种情况下，实现可以为 showmanyc（）提供更有用的实现，但是简单的实现便宜且符合要求。

in_avail() will call showmanyc() and zero (which is the default implementation of this virtual function) is a safe thing to return as it means that a read might block and it might fail but isn't guaranteed to do either. Obviously it is possible for an implementation to provide a more helpful implementation for showmanyc() in this case, but the simple implementation is cheap and conformant.

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