C++ cout << 字符串流
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cout << stringstream
提问by MCP
When I put something into a stringstream, let's say a real number, if I then insert that stringstream object into cout...what am I looking at?
当我将一些东西放入 stringstream 时，假设是一个实数，如果我然后将该 stringstream 对象插入 cout ......我在看什么？
Usually I'm getting some strange number. Is this a memory location? Just curious.
通常我会得到一些奇怪的数字。这是内存位置吗？只是好奇。
It looks like the below comment hit it but here's what I'm trying to do:
看起来下面的评论击中了它，但这是我想要做的：
string stringIn;
stringstream holdBuff;
holdBuff << getline(cin, stringIn);
cout << holdBuff;
Basically I was just trying to see what holdBuff looked like once I inserted stringIn. I am trying to have the user enter a string and then I want to step through it looking for it's contents and possilbly converting...
基本上，我只是想看看当我插入 stringIn 后，holdBuff 是什么样子的。我试图让用户输入一个字符串，然后我想逐步通过它寻找它的内容并可能转换...
回答by James Kanze
What do you think
你觉得怎么样
holdBuff << getline(cin, stringIn);
is doing. The return type of getlineis a reference to the stream
being read (cin) in this case. Since there's no <<defined which
takes an std::istreamas second argument, the compiler tries different
conversions: in C++11, std::istreamhas an implicit conversion to
bool, and in earlier C++, an implicit conversion to std::ios*, or
something similar (but the only valid use of the returned value is to
convert it to bool). So you'll either output 1(C++11), or some
random address (in practice, usually the address of the stream, but this
is not guaranteed). If you want to get the results of a call to
getlineinto an std::ostringstream, you need two operations (with a
check for errors between them):
是在做。在这种情况下，的返回类型getline是对正在读取的流的引用 ( cin)。因为没有<<限定它接受一个std::istream作为第二个参数，编译器试图不同的转换：在C ++ 11，std::istream有一个隐式转换
bool，并且在较早的C ++，一个隐式转换到std::ios*，或类似的东西（但唯一的有效利用的返回值是将其转换为bool)。所以你要么输出1（C++11），要么输出一些随机地址（实际上，通常是流的地址，但这不能保证）。如果你想获得一个调用的结果
getline为std::ostringstream，则需要两个操作（与它们之间的误差校验）：
if ( !getline( std::cin, stringIn ) )
// Error handling here...
holdBuff << stringIn;
Similarly, to write the contents of a std::ostringstream,
类似地，要写入 a 的内容std::ostringstream，
std::cout << holdBuf.str() ;
is the correct solution. If you insist on using an std::stringstreamwhen an std::ostringstreamwould be more appropriate, you can also do:
是正确的解决方案。如果您坚持使用std::stringstreamwhenstd::ostringstream更合适，您还可以这样做：
std::cout << holdBuf.rdbuf();
The first solution is preferable, however, as it is far more idiomatic.
然而，第一种解决方案更可取，因为它更加地道。
In any case, once again, there is no <<operator that takes any
iostreamtype, so you end up with the results of the implicit
conversion to boolor a pointer.
在任何情况下，再一次，没有<<任何iostream类型的运算符
，因此您最终会得到隐式转换为bool或 指针的结果。
回答by Bo Persson
Yes, you are likely to see the address of the stringstream.
是的，您很可能会看到字符串流的地址。
If you want to display the string it contains, try
如果要显示它包含的字符串，请尝试
cout << stream.str();
回答by Goz
Yes it is most likely a memory location of some form or other. Most likely it is the pointer to the stringstream object itself.
是的，它很可能是某种形式的内存位置。它很可能是指向 stringstream 对象本身的指针。
You could confirm this as follows:
您可以通过以下方式确认这一点：
std::stringstream ss;
unsigned long long ll = (unsigned long long)&ss;
cout << ll;
That said when you want to cout a stringstream you should use the str() function as follows:
也就是说，当你想计算一个字符串流时，你应该使用 str() 函数，如下所示：
cout << ss.str();
回答by Michael Krelin - hacker
cout << s.rdbuf();
is what you want. Alternatively you may want to
是你想要的。或者你可能想要
cout << s.str();
which may be more expensive in terms of resources though.
但这在资源方面可能更昂贵。
