C++11 对 lambda 返回类型的限制

Question

我正在阅读 cppreference关于如何推导出 C++11 lambda 的返回类型:

if the body consists of the single return statement, the return type is the type of the returned expression (after rvalue-to-lvalue, array-to-pointer, or function-to-pointer implicit conversion)

所以我认为这意味着一个 lambda 只能有一个 return 语句。但为什么它仍然适用于多个 return 语句？

这compiles在两个编译器上:

auto f = [] (bool c1, bool c2) {
    if (c1) return 1;
    if (c2) return 2;
    else    return 3;
};

Answer 1

这有点不精确。 [expr.prim.lambda]/4:

If a lambda-expression does not include a lambda-declarator, it is as if the lambda-declarator were (). If a lambda-expression does not include a trailing-return-type, it is as if the trailing-return-type denotes the following type:

• if the compound-statement is of the form

  { attribute-specifier-seq_optreturn expression ; }

  the type of the returned expression after lvalue-to-rvalue conversion (4.1), array-to-pointer conversion (4.2), and function-to-pointer conversion (4.3);

• otherwise, void.

因此，只有当 lambda 表达式的整个主体仅由一个单独的 return 语句组成时，才能推导出返回类型。

GCC 和 Clang 都不符合标准这种情况下，当且仅当两个 return 语句导致不一致的推论时，他们才会发出错误消息。这是因为他们已经实现了 C++14 标准，即使存在多个 return 语句和/或多个其他语句，该标准也会扣除返回类型。 [expr.prim.lambda]/4 指定

The lambda return type is auto, which is replaced by the trailing-return-type if provided and/or deduced from return statements as described in 7.1.6.4.

§7.1.6.4/9

If a function with a declared return type that contains a placeholder type has multiple return statements, the return type is deduced for each return statement. If the type deduced is not the same in each deduction, the program is ill-formed.