临时和表达行为
这是明确定义的行为吗？
const char* p = (std::string("Hello") + std::string("World")).c_str();
std::cout << p;
我不知道。原因？
Is this well defined behavior?
const char* p = (std::string("Hello") + std::string("World")).c_str();
std::cout << p;
I am not sure. Reasons?
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不，这是未定义的行为。
std::string临时变量和operator+返回的临时变量都只能存活到const char*初始化结束（完整的结束）表达）。然后它们被销毁并且p指向不确定的内存。No, this is undefined behavior. Both
std::stringtemporaries and the temporary returned byoperator+only live until the end of the initialization of yourconst char*(end of full expression). Then they are destroyed andppoints to uncertain memory.否，行为未定义，因为
p指向std::cout << 中已释放的存储空间。 p;创建一个临时对象来保存
std::string("Hello") + std::string("World")。然后从该对象中检索 C 样式字符串。在表达式末尾，临时对象被销毁，留下p指向已释放的存储空间。然后使用
p调用未定义的行为。12.2/4 说
No the behaviour is undefined because
ppoints to deallocated storage instd::cout << p;A temporary is created to hold
std::string("Hello") + std::string("World"). C-style string is then retrived from that object. At the end of the expression that temporary is destroyed leavingppointing to a deallocated storage.Using
pthen invokes Undefined Behavior.12.2/4 says
由于缺少分号，它不会编译：
现在规则适用，临时值在其使用的表达式末尾（分号处）被删除。所以你有一个指向已删除内存的指针，这会导致未定义的行为。
it won't compile because of a missing semi-colon:
NOW the rule applies that the temporary is deleted at the end of the expression it is used in, which is at the semicolon. So you have a pointer to deleted memory which causes undefined behaviour.
我最近读了这本优秀的书 http://www.amazon.co.uk/Gotchas-Avoiding-Addison-Wesley-Professional-Computing/dp/0321125185/ref 如果我没记错的话，这几乎是给出的示例之一那里。
我相信从
c_str返回的数据仅在返回它的字符串对象有效时才有效。示例中的字符串对象仅在表达式的持续时间内有效。I recently read this excellent book http://www.amazon.co.uk/Gotchas-Avoiding-Addison-Wesley-Professional-Computing/dp/0321125185/ref and if I recall correctly this is pretty much one of the examples given there.
I believe the data returned from
c_stris only valid as long as the string object that returned it is live. The string objects in your example are only live for the duration of the expression.