Effective Modern C++ Item 25(170p~171p)中的代码如下:
class Widget {
public:
void setName(const std::string& newName) // set from
{ name = newName; } // const lvalue
void setName(std::string&& newName) // set from
{ name = std::move(newName); } // rvalue
…
};
w.setName("Adela Novak");
With the version of setName taking a universal reference, the string literal "Adela Novak" would be passed to setName, where it would be conveyed to the assignment operator for the std::string inside w. w’s name data member would thus be assigned directly from the string literal; no temporary std::string objects would arise.
我不明白如果调用采用通用引用的 setName 版本,为什么“不会出现临时 std::string 对象”。不应该将 newName 创建为临时 std::string 吗?
最佳答案
代码前面有
template<typename T> void setName(T&& newName) // newName is { name = std::forward<T>(newName); } // universal reference
稍后出现您复制到问题中的代码。然后,在您引用的文字之前:
That would certainly work in this case, but there are drawbacks. [...] For example, consider this use of
setName
:w.setName("Adela Novak");
紧接着:
With the overloaded versions of
setName
, however, [...]
文本是说除了您引用的代码之外,其他代码中不会出现临时字符串对象。您严重误读并因此错误引用了文本。
关于c++ - 为什么 Effective Modern C++ 的第 25 项中说 "no temporary std::string objects would arise"?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48368673/