c++ - 开关盒 : declaration-with-initialization & declaration-and-then-assignment

Question

在 switch-case 语句中，declaration-with-initialization 是无效的，但允许 declaration-and-then-assignment。如以下代码片段所示。

从编译器端看，这两种类型的初始化有什么区别？以及为什么第一种初始化无效而第二种初始化有效。

switch(val)  
{  
case 0:  
  int newVal = 42;  //Invalid
  break;
case 1:  
  int newVal2;      //Valid
  newVal2 = 42;  
  break;
case 2:
  break;
}

Answer 1

实际上，规则是您不能跳入经过具有初始化的声明(或经过非 POD 类型变量的声明)的 block 。 C++ 标准说 (C++03 §6.7):

It is possible to transfer into a block, but not in a way that bypasses declarations with initialization. A program that jumps⁽⁷⁷⁾ from a point where a local variable with automatic storage duration is not in scope to a point where it is in scope is ill-formed unless the variable has POD type (3.9) and is declared without an initializer (8.5).

^{(*) The transfer from the condition of a switch statement to a case label is considered a jump in this respect.}

int newVal = 42; 是一个带有初始值设定项的声明(= 42 部分)。该程序格式错误，因为如果 val 是 1 或 2，您将在初始化后跳转到 switch block 。

int newVal2;也是一个声明；因为 int 是一个 POD 类型并且声明没有初始化器，你可以跳过这个声明。