试试这个:

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

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;
}

c++标准有: 可以将其转移到块中，但不能绕过带有初始化的声明。如果一个程序从一个具有自动存储持续时间的局部变量不在作用域中的点跳转到它在作用域中的点，那么该程序就是病态形式的，除非该变量具有POD类型(3.9)，并且声明时没有初始化式(8.5)。

说明此规则的代码:

#include <iostream>

using namespace std;

class X {
  public:
    X() 
    {
     cout << "constructor" << endl;
    }
    ~X() 
    {
     cout << "destructor" << endl;
    }
};

template <class type>
void ill_formed()
{
  goto lx;
ly:
  type a;
lx:
  goto ly;
}

template <class type>
void ok()
{
ly:
  type a;
lx:
  goto ly;
}

void test_class()
{
  ok<X>();
  // compile error
  ill_formed<X>();
}

void test_scalar() 
{
  ok<int>();
  ill_formed<int>();
}

int main(int argc, const char *argv[]) 
{
  return 0;
}

显示初始化器效果的代码:

#include <iostream>

using namespace std;

int test1()
{
  int i = 0;
  // There jumps fo "case 1" and "case 2"
  switch(i) {
    case 1:
      // Compile error because of the initializer
      int r = 1; 
      break;
    case 2:
      break;
  };
}

void test2()
{
  int i = 2;
  switch(i) {
    case 1:
      int r;
      r= 1; 
      break;
    case 2:
      cout << "r: " << r << endl;
      break;
  };
}

int main(int argc, const char *argv[]) 
{
  test1();
  test2();
  return 0;
}

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。

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。

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。