参见英文答案 > Are parentheses around the result significant in a return statement?                                    12个
情况1：

#include <iostream>

decltype(auto) fun()
{
        std::string str = "In fun";
        return str;
}

int main()
{
        std::cout << fun() << std::endl;
}

在这里,程序在Gcc编译器中工作正常. decltype(auto)被推断为str的类型.

案例2：

#include <iostream>

decltype(auto) fun()
{
        std::string str = "In fun";
        return (str); // Why not working??
}

int main()
{
        std::cout << fun() << std::endl;
}

在这里,生成以下错误和分段错误：

In function 'decltype(auto) fun()':
prog.cc:5:21: warning: reference to local variable 'str' returned [-Wreturn-local-addr]
         std::string str = "In fun";
                     ^~~
Segmentation fault

为什么返回(str);给出分段错误？

解决方法:

decltype以两种不同的方式工作;当使用带有未表示的id-expression时,它会产生它所声明的确切类型(在第一种情况下它是std :: string).除此以外,

If the argument is any other expression of type T, and

a) if the value category of expression is xvalue, then decltype yields
T&&;

b) if the value category of expression is lvalue, then decltype yields
T&;

c) if the value category of expression is prvalue, then decltype
yields T.

和

Note that if the name of an object is parenthesized, it is treated as an ordinary lvalue expression, thus decltype(x) and decltype((x)) are often different types.

(str)是带括号的表达式,它是一个左值;然后它产生字符串&的类型.因此,您将返回对局部变量的引用,它将始终悬空.取消引用它会导致UB.