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Switch-Case: declaration-with-initialization & declaration-and-then-assignment

In the switch-case statements declaration-with-initialization is invalid but declaration-and-then-assignment is allowed. As shown in the following code snippet.

What is difference between these two type of initializations from the compiler side? And why is the first type of initialization invalid and second type a valid one.

switch(val)  
{  
case 0:  
  int newVal = 42;  //Invalid
  break;
case 1:  
  int newVal2;      //Valid
  newVal2 = 42;  
  break;
case 2:
  break;
}
like image 284
Manjeet Dahiya Avatar asked Sep 21 '10 05:09

Manjeet Dahiya


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1 Answers

Effectively, the rule is that you can't jump into a block past a declaration that has an initialization (or past the declaration of a non-POD type variable). The C++ standard says (C++03 §6.7):

It is possible to transfer into a block, but not in a way that bypasses declarations with initialization. A program that jumps(77) 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 transfer from the condition of a switch statement to a case label is considered a jump in this respect.

int newVal = 42; is a declaration that has an initializer (the = 42 part). The program is ill-formed because if val is 1 or 2, you'll jump into the switch block past the initialization.

int newVal2; is also a declaration; because int is a POD type and the declaration has no initializer, you can jump past this declaration.

like image 161
James McNellis Avatar answered Sep 22 '22 01:09

James McNellis