Case-Label question in C/C++

Question

I came across this code

  #include<stdio.h>
  int main()
  {
      int a=1;
      switch(a)
      {   int b=20;
          case 1: printf("b is %d\n",b);
                  break;
          default:printf("b is %d\n",b);
                  break;
      }
      return 0;
  }

I expected the output to be 20 but got some garbage value. Will the output be different when this code is compiled as a .c file and .cpp file?

Answer 1

In C++ the code is ill-formed because jump to case 1 crosses initialization of b. You can't do that.

In C the code invokes UB because of the use of uninitialized variable b.

C99 [6.4.2/7] also shows a similar example.

EXAMPLE In the artificial program fragment

switch (expr)
{
   int i = 4;
   f(i);
   case 0:
     i = 17;
     /* falls through into defaultcode  */
   default:
     printf("%d\n", i);
}

the object whose identifier is i exists with automatic storage duration (within the block) but is never initialized, and thus if the controlling expression has a nonzero value, the call to the printf function will access an indeterminate value. Similarly, the call to the function fcannot be reached.