如何理解"well formed when treated as an unevaluated operand"

Question

根据 document，它说[强调我的]：

template <class Fn, class... ArgTypes>
struct is_invocable; 
Determines whether Fn can be invoked with the arguments ArgTypes.... Formally, determines whether INVOKE(declval<Fn>(), declval<ArgTypes>()...) is well formed when treated as an unevaluated operand, where INVOKE is the operation defined in Callable.

如何理解加粗的语句？什么是“未计算的操作数”？

也许，一个简单的例子有助于充分理解这个问题。

Answer 1

How to understand the statement in bold? What's "an unevaluated operand"?

未计算的操作数是未计算的操作数。

Maybe, a simple example helps to fully understand this matter.

void g(auto) requires false;
void f(auto x) { g(x); }

static_assert(is_invocable_v<decltype(f<int>), int>);
f(0); // ill-formed

考虑到上面的例子，由于is_invocable没有计算，f中的g(x)实际上也不会被调用，所以static_assert会通过。

由于f(0)实际被求值，会因为g(0)不满足约束条件而失败

如何理解"well formed when treated as an unevaluated operand"

How to understand "well formed when treated as an unevaluated operand"

c++

c++17

callable