C++ 11自动和函数返回类型

Question

我知道之间的区别的auto,auto&,const auto和const auto&(例如在"每个"循环),但有一两件事令我惊讶的是:

std::string bla;
const std::string& cf()
{
    return bla;
}


int main (int argc, char *argv[])
{
    auto s1=cf();
    const std::string& s2=cf();
    s1+="XXX"; // not an error
    s2+="YYY"; //error as expected
}

那么有人可以告诉我何时x表达式auto x = fun();中的类型与返回值的类型不是同一类型fun()？

Answer 1

规则auto与模板类型推导相同:

template <typename T>
void f(T t); // same as auto
template <typename T>
void g(T& t); // same as auto&
template <typename T>
void h(T&& t); // same as auto&&

std::string sv;
std::string& sl = sv;
std::string const& scl = sv;

f(sv); // deduces T=std::string
f(sl); // deduces T=std::string
f(scl); // deduces T=std::string
f(std::string()); // deduces T=std::string
f(std::move(sv)); // deduces T=std::string

g(sv); // deduces T=std::string, T& becomes std::string&
g(sl); // deduces T=std::string, T& becomes std::string&
g(scl); // deduces T=std::string const, T& becomes std::string const&
g(std::string()); // does not compile
g(std::move(sv)); // does not compile

h(sv); // deduces std::string&, T&& becomes std::string&
h(sl); // deduces std::string&, T&& becomes std::string&
h(scl); // deduces std::string const&, T&& becomes std::string const&
h(std::string()); // deduces std::string, T&& becomes std::string&&
h(std::move(sv)); // deduces std::string, T&& becomes std::string&&

通常,如果您需要副本,则使用auto,如果您需要使用的引用auto&&.auto&&保留referene的常量,也可以绑定临时(延长其寿命).