尾随返回类型的占位符是否会覆盖初始占位符? [英] Does a placeholder in a trailing-return-type override an initial placeholder?

问题描述

g ++似乎接受auto和decltype(auto)的任何组合作为初始和尾随返回类型:

g++ appears to accept any combination of auto and decltype(auto) as initial and trailing return types:

int a;
auto f() { return (a); }                             // int
auto g() -> auto { return (a); }                     // int
auto h() -> decltype(auto) { return (a); }           // int&
decltype(auto) i() { return (a); }                   // int&
decltype(auto) j() -> auto { return (a); }           // int
decltype(auto) k() -> decltype(auto) { return (a); } // int&

但是，clang拒绝j和k，说:错误:尾随返回类型的函数必须指定返回类型'auto'，而不是'decltype(auto)'(演示).

However, clang rejects j and k, saying: error: function with trailing return type must specify return type 'auto', not 'decltype(auto)' (demonstration).

哪个编译器正确?在每种情况下应使用哪个规则(auto或decltype(auto))?在 trailing-return-type 中使用占位符类型有意义吗?

Which compiler is correct? Which rule (auto or decltype(auto)) should be used in each case? And does it make any sense to use a placeholder type in a trailing-return-type?

推荐答案

auto.

§8.3.5[dcl.fct]/2

§8.3.5 [dcl.fct] /2

在声明T D中，其中D具有形式

In a declaration T D where D has the form

D1 ( parameter-declaration-clause ) cv-qualifier-seq _opt ref-qualifier _opt exception-specification _opt attribute-specifier-seq _opt trailing-return-type

D1 ( parameter-declaration-clause ) cv-qualifier-seq_opt ref-qualifier_opt exception-specification_opt attribute-specifier-seq_opt trailing-return-type

，声明T D1中包含的 declarator-id 的类型为"derived- declarator-type-list T"，

and the type of the contained declarator-id in the declaration T D1 is "derived-declarator-type-list T",

T应该是单个 type-specifier 自动. [...]

T shall be the single type-specifier auto. [...]

另请参阅核心问题1852 与[dcl.spec.auto]/1矛盾.

See also Core Issue 1852 for the apparent contradiction with [dcl.spec.auto]/1.

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