到目前为止，大多数回复在一个方面是错误的:你可以在case语句之后声明变量，但你不能初始化它们:

case 1:
    int x; // Works
    int y = 0; // Error, initialization is skipped by case
    break;
case 2:
    ...

如前所述，解决这个问题的一个好方法是使用大括号为案例创建作用域。

考虑:

switch(val)
{
case VAL:
   int newVal = 42;
default:
   int newVal = 23;
}

在没有break语句的情况下，有时newVal会声明两次，直到运行时才知道它是否声明了。我的猜测是，这种限制是由于这种混乱。newVal的范围是什么?惯例规定它将是整个开关块(在大括号之间)。

我不是c++程序员，但在C中:

switch(val) {
    int x;
    case VAL:
        x=1;
}

工作很好。在开关块中声明变量是可以的。在案件警卫之后宣布不是。

如果你的代码说“int newVal=42”，那么你可以合理地期望newVal永远不会是未初始化的。但是如果你浏览这条语句(这就是你正在做的)，那么就会发生这样的事情——newVal在作用域内，但没有被分配。

如果这是你真正想要发生的，那么语言需要通过“int newVal;newVal = 42;"。否则，您可以将newVal的范围限制为单个情况，这更可能是您想要的结果。

如果你考虑相同的例子，但加上"const int newVal = 42;"

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。

newVal存在于开关的整个作用域中，但只有在VAL分支被命中时才会初始化。如果你在VAL中创建一个代码块，它应该是OK的。

一个switch块不同于一连串的if/else if块。我很惊讶没有其他答案能解释清楚。

考虑下面的switch语句:

switch (value) {
    case 1:
        int a = 10;
        break;
    case 2:
        int a = 20;
        break;
}

这可能令人惊讶，但编译器不会将其视为简单的if/else if。它将生成以下代码:

if (value == 1)
    goto label_1;
else if (value == 2)
    goto label_2;
else
    goto label_end;

{
label_1:
    int a = 10;
    goto label_end;
label_2:
    int a = 20; // Already declared !
    goto label_end;
}

label_end:
    // The code after the switch block

case语句被转换为标签，然后用goto调用。括号创建了一个新的作用域，现在很容易看出为什么不能在一个开关块中声明两个具有相同名称的变量。

它可能看起来很奇怪，但是支持fallthrough是必要的(也就是说，不使用break让执行继续到下一个case)。