打印 Stringstream 输出指针
下面的代码没有输出预期的“Bar”字符串，而是输出看起来像指针的内容。
#include <sstream>
#include <iostream>
int main()
{
std::stringstream Foo("Bar");
std::cout << Foo << std::endl; // Outputs "0x22fe80."
return 0;
}
这可以通过使用 Foo.str() 来解决，但它过去给我带来了一些问题。是什么导致了这种奇怪的行为？它记录在哪里？
Rather than outputting the expected string of "Bar", the following code outputs what looks to be a pointer.
#include <sstream>
#include <iostream>
int main()
{
std::stringstream Foo("Bar");
std::cout << Foo << std::endl; // Outputs "0x22fe80."
return 0;
}
This can be worked around by using Foo.str(), but it's caused some issues for me in the past. What causes this strange behavior? Where is it documented?
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评论（3）
这一点也不奇怪。
流就是流，不是字符串。
可以通过成员函数
.str()获取[string]缓冲区。这不是一个解决方法：它是 API。它应该记录在您的 C++ 标准库参考中，如果没有，那么您的参考不完整/错误。（稍微奇怪的是，您得到的是内存地址而不是编译错误；这是由于在 pre-C++ 中使用了safe bool idiom 0x 标准库，如另一个答案中所述。）
It's not strange at all.
A stream is a stream, not a string.
You can obtain the [string] buffer with the member function
.str(). That's not a workaround: it's the API. It should be documented in your C++ standard library reference, and if it's not then your reference is incomplete/wrong.(What is slightly strange is that you get a memory address rather than a compilation error; that is due to the use of the safe bool idiom in the pre-C++0x standard library, as covered in another answer.)
std::stringstream不应插入流中。这就是为什么你有str()方法。您观察到的行为是由于
std::stringstream转换为void*，这是用于测试流的：在 C+ 中+11，此转换被显式转换为 bool 所取代，这导致此代码无法编译：
A
std::stringstreamis not supposed to be inserted in a stream. That's why you have thestr()method.The behavior you're observing is due to the fact that
std::stringstreamhas a conversion tovoid*, which is what is used to test the stream:In C++11, this conversion is replaced by an explicit conversion to bool, which causes this code not to compile:
(1)使用 std::cout <<富<< std::endl;您应该确保 stringstream 已经重载“<<”。
(2)如果没有过载“<<” ，使用 std::cout <<富<< std::endl;可以输出 Foo 的地址。
(1)use std::cout << Foo << std::endl; you should make sure stringstream has already overload "<<".
(2)if there is not overload "<<" , use std::cout << Foo << std::endl; may output the Foo's address.