当不需要检查缩小时，是否需要实例化函数定义？

Question

考虑以下程序：

template<typename T>
constexpr int f() 
{
    T{}.i; // error if instantiated with [T = double]
    return 42;
}

constexpr void g(char);

using U = decltype( g( {f<double>()} ) );

根据我的理解，最后一行是一个错误，因为对 f<double>() 的调用是在大括号初始值设定项中，即使 f<T> returns 和 int，需要返回 int 的值来决定是否可以将其缩小为 g 所期望的 char。这需要用 double 实例化 f 的定义，这会导致错误。 gcc 和 clang 都拒绝此代码。

但是，如果 g 的定义更改为接受 int 参数：

constexpr void g(int);

那么好像不需要实例化f的定义，因为缩小转换必须成功。事实上，gcc 接受这个，但 clang 仍然用 double 实例化 f 并拒绝 code。此外，如果仅声明 f 而未定义，则 clang 接受代码，这意味着不需要定义，也不应实例化。

我的推理是否正确，这是一个 clang 错误，还是需要实例化，这实际上是一个 gcc 错误？

Answer 1

这是CWG #1581 I think, resolved by P0859。

temp.inst/5 说：

Unless a function template specialization is a declared specialization, the function template specialization is implicitly instantiated when the specialization is referenced in a context that requires a function definition to exist or if the existence of the definition affects the semantics of the program.

存在是否影响程序的语义？

temp.inst/8 说：

The existence of a definition of a variable or function is considered to affect the semantics of the program if the variable or function is needed for constant evaluation by an expression ([expr.const]), even if constant evaluation of the expression is not required or if constant expression evaluation does not use the definition.

是否需要通过表达式进行常量求值？

expr.const/15.6-7 说：

A function or variable is needed for constant evaluation if it is:

a constexpr function that is named by an expression that is potentially constant evaluated, or

a variable whose name appears as a potentially constant evaluated expression that is either a constexpr variable or is of non-volatile const-qualified integral type or of reference type.

它是由一个可能被评估为常量的表达式命名的吗？

expr.const/15.1-15.5 说：

An expression or conversion is potentially constant evaluated if it is:

a manifestly constant-evaluated expression,

a potentially-evaluated expression,

an immediate subexpression of a braced-init-list,

an expression of the form & cast-expression that occurs within a templated entity, or

a subexpression of one of the above that is not a subexpression of a nested unevaluated operand.

它是 braced-init-list 的直接子表达式。

当不需要检查缩小时，是否需要实例化函数定义？

Is a function definition required to be instantiated when there is no need to check for narrowing?

c++

language-lawyer

c++20

narrowing