这个问题最初同时被标记为c和c++。原始代码在C和c++中都是无效的，但原因完全不同，互不相关。

In C++ this code is invalid because the case ANOTHER_VAL: label jumps into the scope of variable newVal bypassing its initialization. Jumps that bypass initialization of automatic objects are illegal in C++. This side of the issue is correctly addressed by most answers. However, in C language bypassing variable initialization is not an error. Jumping into the scope of a variable over its initialization is legal in C. It simply means that the variable is left uninitialized. The original code does not compile in C for a completely different reason. Label case VAL: in the original code is attached to the declaration of variable newVal. In C language declarations are not statements. They cannot be labeled. And this is what causes the error when this code is interpreted as C code. switch (val) { case VAL: /* <- C error is here */ int newVal = 42; break; case ANOTHER_VAL: /* <- C++ error is here */ ... break; } Adding an extra {} block fixes both C++ and C problems, even though these problems happen to be very different. On the C++ side it restricts the scope of newVal, making sure that case ANOTHER_VAL: no longer jumps into that scope, which eliminates the C++ issue. On the C side that extra {} introduces a compound statement, thus making the case VAL: label to apply to a statement, which eliminates the C issue. In C case the problem can be easily solved without the {}. Just add an empty statement after the case VAL: label and the code will become valid switch (val) { case VAL:; /* Now it works in C! */ int newVal = 42; break; case ANOTHER_VAL: ... break; } Note that even though it is now valid from C point of view, it remains invalid from C++ point of view. Symmetrically, in C++ case the the problem can be easily solved without the {}. Just remove the initializer from variable declaration and the code will become valid switch (val) { case VAL: int newVal; newVal = 42; break; case ANOTHER_VAL: /* Now it works in C++! */ ... break; } Note that even though it is now valid from C++ point of view, it remains invalid from C point of view.

从C23开始，C语言中的所有标签都将被解释为标签隐含的空语句(N2508)，也就是说，在C语言中不能将标签放在声明前面的问题将不再存在，并且不再需要上述基于;的修复。

到目前为止，答案都是c++。

对于c++，你不能跳过初始化。但是，在C语言中，声明不是语句，大小写标签后面必须跟着语句。

所以，有效(但丑陋)的C，无效的c++

switch (something)
{
  case 1:; // Ugly hack empty statement
    int i = 6;
    do_stuff_with_i(i);
    break;
  case 2:
    do_something();
    break;
  default:
    get_a_life();
}

相反，在c++中，声明是一个语句，因此下面的语句是有效的c++，无效的C

switch (something)
{
  case 1:
    do_something();
    break;
  case 2:
    int i = 12;
    do_something_else();
}

整个switch语句在同一个作用域中。要解决这个问题，可以这样做:

switch (val)
{
    case VAL:
    {
        // This **will** work
        int newVal = 42;
    }
    break;

    case ANOTHER_VAL:
      ...
    break;
}

注意括号。

试试这个:

switch (val)
{
    case VAL:
    {
        int newVal = 42;
    }
    break;
}

我最喜欢的邪恶切换技巧是使用if(0)跳过不需要的case标签。

switch(val)
{
case 0:
// Do something
if (0) {
case 1:
// Do something else
}
case 2:
// Do something in all cases
}

但非常邪恶。

我只是想强调斯利姆的观点。转换构造创建了一个完整的、一等公民的范围。因此，可以在switch语句中在第一个case标签之前声明(并初始化)一个变量，而不需要额外的括号对:

switch (val) {  
  /* This *will* work, even in C89 */
  int newVal = 42;  
case VAL:
  newVal = 1984; 
  break;
case ANOTHER_VAL:  
  newVal = 2001;
  break;
}