initialization in switch

Hello all

consider the code fragment below:
case 1:
{
int i = 1;
break;
}

VC give this error:error C2360: initialization of 'i' is skipped by
'case' label

but if i write like this:
case 1:
{
int i;
i = 1;
break;
}

it is successfully compiled


why ?

re: initialization in switch

"Teddy" <duanduan12@hotmail.com> wrote in message
news:1117416598.164487.151050@g44g2000cwa.googlegr oups.com...[color=blue]
> Hello all
>
> consider the code fragment below:
> case 1:
> {
> int i = 1;
> break;
> }
>
> VC give this error:error C2360: initialization of 'i' is skipped by
> 'case' label
>
> but if i write like this:
> case 1:
> {
> int i;
> i = 1;
> break;
> }
>
> it is successfully compiled
>
>
> why ?[/color]

Well, what happens when somewhere down the line, after the switch, or even
in a differant case, the value of i is queried? e.g. x = i; Unless case 1 is
guarenteed then there is no such thing as i. Because case 1 is the only
block aware of i's existance at all. In the second code snipper there is
always an i. The compiler is smart enough to know this.

re: initialization in switch

On 2005-05-30, Teddy <duanduan12@hotmail.com> wrote:[color=blue]
> Hello all
>
> consider the code fragment below:
> case 1:
> {
> int i = 1;
> break;
> }[/color]

Are you sure you did exactly the above, and not something like this:

switch (i) {
case 1:
int i = 1;
break;
case 2: // skips initialization of i, but i still has scope!
// is the variable i initialized or not ? it's in limbo state

This should give such an error, but the code you posted shouldn't.
[color=blue]
> VC give this error:error C2360: initialization of 'i' is skipped by
> 'case' label
>
> but if i write like this:
> case 1:
> {
> int i;
> i = 1;
> break;
> }
>
> it is successfully compiled[/color]

Because i is not initialized.

Cheers,
--
Donovan Rebbechi
http://pegasus.rutgers.edu/~elflord/

re: initialization in switch

On 2005-05-30, Christopher <anon@nospam.net> wrote:
[color=blue]
> Well, what happens when somewhere down the line, after the switch, or even
> in a differant case, the value of i is queried? e.g. x = i;[/color]

In the code he posted, i has block scope. So there's no ambiguity -- if that
happens, it's an "undeclared identifier".
[color=blue]
> Unless case 1 is
> guarenteed then there is no such thing as i.[/color]

What about this :

if (flag == 0) {
int i = 0;
} else if ( flag == 1) {
int j = 0;
} else if ( flag == 2) {
int k = 0;
}

if flag is set to 0, then there is no such thing as i either.
[color=blue]
> Because case 1 is the only
> block aware of i's existance at all.[/color]

Likewise above: only the first block knows about i. Yet it is legal.
[color=blue]
> In the second code snipper there is
> always an i. The compiler is smart enough to know this.[/color]

Are you saying that block scope is only allowed if the compiler knows
whether or not that block is executed ?

Cheers,
--
Donovan Rebbechi
http://pegasus.rutgers.edu/~elflord/

re: initialization in switch

"Christopher" <anon@nospam.net> wrote in message
news:mIume.23756$6g3.5385@tornado.texas.rr.com...[color=blue]
>
> "Teddy" <duanduan12@hotmail.com> wrote in message
> news:1117416598.164487.151050@g44g2000cwa.googlegr oups.com...[color=green]
> > Hello all
> >
> > consider the code fragment below:
> > case 1:
> > {
> > int i = 1;
> > break;
> > }
> >
> > VC give this error:error C2360: initialization of 'i' is skipped by
> > 'case' label
> >
> > but if i write like this:
> > case 1:
> > {
> > int i;
> > i = 1;
> > break;
> > }
> >
> > it is successfully compiled
> >
> >
> > why ?[/color][/color]

Using what you gave as example, I compiled and excuted successfully with
VC++ 6.0 and MinGW 3.4.2
You will need to give fully failing program in order for checking it here.

#include <iostream>
#include <ostream>

int main( )
{
int j = 1;
switch (j)
{
case 1:
{
int i = 1;
break;
}
}
std::cout << j << std::endl;
return 0;
}

--
Gary

re: initialization in switch

It was my fault
I posted the wrong code