当规则说初始化的完整表达式时，这是完整表达式

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full-expression的定义：

A full-expression is

an unevaluated operand,

a constant-expression,

an init-declarator or a mem-initializer, including the constituent expressions of the initializer,

an invocation of a destructor generated at the end of the lifetime of an object other than a temporary object, or

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.

项目符号 3 说 s1 = 1 是一个完整表达式，因为它是一个初始化声明符，而 I(i) 是一个完整表达式，因为它是一个 mem-initializer 并且对于 D(i)。这意味着初始化实体 s1 可以包含多个完整表达式？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，完整表达式可以动态嵌套：考虑

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

因此，初始化 s1 是 init-declarator 完整表达式的一部分，同时还包含其 [=16= 的完整表达式之间没有冲突]内存初始化器s.

当规则说初始化的完整表达式时，这是完整表达式

which is the full-expression when the rule says the full-expression of initialization

c++

language-lawyer

c++17