c++ - 当规则说初始化的完整表达式时，它是完整表达式

Question

struct S {
  constexpr S(int i): I(i),D(i) { }       // full-expressions are initialization of I and initialization of D 
private:
  int I;
  int D;
};
int main(){
  constexpr S s1 = 1;  //full-expression comprises call of S​::​S(int)
}

根据全表达式的定义:

A full-expression is

1. an unevaluated operand,
2. a constant-expression,
3. an init-declarator or a mem-initializer, including the constituent expressions of the initializer,
4. an invocation of a destructor generated at the end of the lifetime of an object other than a temporary object, or
5. an expression that is not a subexpression of another expression and that is not otherwise part of a full-expression.

For an initializer, performing the initialization of the entity (including evaluating default member initializers of an aggregate) is also considered part of the full-expression.

子弹 3 说 s1 = 1是一个完整的表达式，因为它是一个初始化声明符和 I(i)是一个完整的表达式，因为它是 mem-initializer和 D(i) 类似.表示初始化实体 s1可以包含多个完整表达式吗？在这种情况下，这组全表达式中初始化的全表达式是哪个？

Answer 1

当然全表达式可以是动态 嵌套:考虑

void f(int i) {
  ++i;   // (useless) full-expression
}
void g() {
  f(1);  // full-expression
}

因此，初始化 s1 之间没有冲突。作为 init-declarator full-expression 的一部分，同时还包含其 mem-initializers 的完整表达式。