c – 为什么没有初始化器跳进标量类型对象的范围？

当我正在阅读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.

它为什么要起作用？跳过它的定义并使用undefined x不是很危险吗？为什么初始化器的存在会有什么不同？

解决方法

无论如何你都会使用未初始化的x,因为int x;就像它会得到的那样没有初始化.初始化程序的存在当然是有区别的,因为你要跳过它. int x = 5;例如,初始化x,所以如果你跳过它就会有所不同.

原文地址：https://www.jb51.cc/c/110507.html