从ildjarn阅读this answer后,我写了下面的例子,它看起来像一个无名的临时对象具有相同的续航时间作为参考!引用一位不愿透露姓名的临时对象(生命周期)
- 这怎么可能?
- 是否在C++标准中指定?
- 哪个版本?
源代码:
#include <iostream> //cout
#include <sstream> //ostringstream
int main()
{
std::ostringstream oss;
oss << 1234;
std::string const& str = oss.str();
char const* ptr = str.c_str();
// Change the stream content
oss << "_more_stuff_";
oss.str(""); //reset
oss << "Beginning";
std::cout << oss.str() <<'\n';
// Fill the call stack
// ... create many local variables, call functions...
// Change again the stream content
oss << "Again";
oss.str(""); //reset
oss << "Next should be '1234': ";
std::cout << oss.str() <<'\n';
// Check if the ptr is still unchanged
std::cout << ptr << std::endl;
}
执行:
> g++ --version
g++ (GCC) 4.1.2 20080704 (Red Hat 4.1.2-54)
Copyright (C) 2006 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
> g++ main.cpp -O3
> ./a.out
Beginning
Next should be '1234':
1234
“const”引用延长了临时对象的生命周期的事实应该在你的C++书中相当公开。 – 2013-03-07 09:41:48
如果您只是尝试在网页上搜索您使用的短语,您应该可以找到答案:_临时对象与其reference_具有相同的生存时间_ – 2013-03-07 09:48:16
[临时生命期]的可能重复(http://stackoverflow.com/questions/4214153/lifetime-of-temporaries)/看看我发现了什么 – 2013-03-07 09:54:45