Why can't variables be declared in a switch statement?

Question

I\'ve always wondered this - why can\'t you declare variables after a case label in a switch statement? In C++ you can declare variables pretty much anywhere (and declaring

Answer 1

C++ Standard has: 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 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 code to illustrate this rule:

#include 

using namespace std;

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

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

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

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

void test_scalar() 
{
  ok();
  ill_formed();
}

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

The code to show the initializer effect:

#include 

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;
}