c++ - 为什么可以跳入不带初始值设定项的标量类型对象的范围？

Question

当我阅读 C++ 标准时，根据标准，以下代码似乎完全没问题。

int main() {
   goto lol;
   {
      int x;
lol:
      cout << x << endl;
   }
}

// OK

[n3290: 6.7/3]: It is possible to transfer into a block, but not in a way that bypasses declarations with initialization. A program that jumps from a point where a variable with automatic storage duration is not in scope to a point where it is in scope is ill-formed unless the variable has scalar type, class type with a trivial default constructor and a trivial destructor, a cv-qualified version of one of these types, or an array of one of the preceding types and is declared without an initializer.

为什么它会起作用？跳过它的定义并使用未定义的 x 不是仍然很危险吗？为什么初始化程序的存在会有所不同？

Answer 1

无论如何，您都会使用未初始化的 x，因为 int x; 是未初始化的。初始化程序的存在当然会有所不同，因为您会跳过它。 int x = 5; 例如初始化 x，所以如果你跳过它会有所不同。