This case bypasses initialization of a local variable

[gyan]

follwing code gives error:
1 #include<iostream.h>
2 int main()
3 {
4 int a=5,b;
5 switch(a){
6 case 1:
7 {b=5;
8 break;
9 }
10 case 2:
11 b=7;
12 int c=4;
13 c=3;
14 break;
15 case 3:
16 c=2;
17 break;
18 }
19 return 0;
20 }


"1.cpp", line 15: Error: This case bypasses initialization of a local
variable.

If i put statements under case 2: in {}, i get error:
"1.cpp", line 18: Error: c is not defined.

So exactly what is happening?
i have read that
A problem occurs , when a variable is declared AND INITIALIZED, in a
location where the flow-of-control is undetermined relative to the
location of the declaration (after the declaration has been moved to the
beginning of the scope block).

Or perhaps a better way of saying it, is the compiler recognizes the
declaration and initialization as two different things:

the declaration is moved to a "static" location at the beginning of the
scope block
while the initialization is really an assignment statement whose execution
is "dynamic", that is, it depends on the flow of control through the
program


Can someone one explain reason in a more common way. I am not able to
understand it.Can you put code, as seen by compiler.

 

[Victor Bazarov]

follwing code gives error:
1 #include<iostream.h>
2 int main()
3 {
4 int a=5,b;
5 switch(a){
6 case 1:
7 {b=5;
8 break;
9 }
10 case 2:
11 b=7;
12 int c=4;
13 c=3;
14 break;
15 case 3:
16 c=2;

'c' is local to this 'switch'. However, if 'a' is 3, then the declaration
of 'c' (and its initialisation) does not get "executed". That's forbidden.

17 break;
18 }
19 return 0;
20 }


"1.cpp", line 15: Error: This case bypasses initialization of a local
variable.

If i put statements under case 2: in {}, i get error:
"1.cpp", line 18: Error: c is not defined.

So exactly what is happening?

What's happening is simple: you're trying to declare/define/initialise
a variable in one place and use it later, but the control flow _allows_
the delcaration/definition/initialisation to be skipped. Similar to

goto blah;
int c = 5; // this is skipped by the goto.
blah:
c = 2;

i have read that
A problem occurs , when a variable is declared AND INITIALIZED, in a
location where the flow-of-control is undetermined relative to the
location of the declaration (after the declaration has been moved to
the beginning of the scope block).

I am not sure I understand that (my coffee hasn't kicked in yet).

Or perhaps a better way of saying it, is the compiler recognizes the
declaration and initialization as two different things:

the declaration is moved to a "static" location at the beginning of
the scope block
while the initialization is really an assignment statement whose
execution is "dynamic", that is, it depends on the flow of control
through the program


Can someone one explain reason in a more common way. I am not able to
understand it.Can you put code, as seen by compiler.

The relevant part of your switch statement is semantically equivalent
to this:

if (a == 3) goto case3;
case2:
int c = 4;
c = 3;
case3:
c = 2;

If 'a' is indeed 3, the code that declares/defines/initialises 'c' is
stepped over.

V

 

[Kyle]

gyan napisa³(a):

follwing code gives error:
1 #include<iostream.h>

2 int main()
3 {
4 int a=5,b;
5 switch(a){
6 case 1:
7 {b=5;
8 break;
9 }
10 case 2:
11 b=7;
12 int c=4;
13 c=3;
14 break;
15 case 3:
16 c=2;
17 break;
18 }
19 return 0;
20 }

please, do not include line numbers like that, it prevents us from
copy-pasting code form newsreader in order to compile it

instead try indicating interesting lines in a comment:


int c=4;
c=3;
break;
case 3: //line 15
c=2;
break;
}

"1.cpp", line 15: Error: This case bypasses initialization of a local
variable.

If i put statements under case 2: in {}, i get error:
"1.cpp", line 18: Error: c is not defined.

So exactly what is happening?

its pretty strightforward, 'c' begins its life at line 12, it is however
visible in 'case 3' becouse its in the same scope,
when you enter 'case 3' you try to use variable 'c', but you ommit its
birth place/time which result in your first error message

when you intorduce a scope in 'case 2' 'c' is not visible in 'case 3'
thus second error message

 

[Christopher Benson-Manica]

The relevant part of your switch statement is semantically equivalent
to this:

if (a == 3) goto case3;
case2:
int c = 4;
c = 3;
case3:
c = 2;

If it is not provable that a is always 3, must a conforming compiler
accept this code (possibly with helpful warnings)?

 

[gyan]

So basically all of you saying that i am declaring/defining/initializing c
in case:2 and using it in case 3. Now it may possible that case 2 block
didn't got executed and in case 3: we are using variable c without
defining it.
But look if i put concern part of code as below:
int a=3;
switch(a){
case 2:
int c;
break;
case 3:
c=5;
cout<<c<<end;
}

I got it compiled and executed successfully.Why?

 

[gyan]

Hi All,
I will like to make my question more straight forward. See code as
#include<iostream.h>
int main()
{
int a=1;
switch(a){
case 1:
int b=3;
break;
case 2:
break;

}
return 0;
}

When compiling, i get error:
"1.cpp", line 9: Error: This case bypasses initialization of a local
variable.

 

[Victor Bazarov]

If it is not provable that a is always 3, must a conforming compiler
accept this code (possibly with helpful warnings)?

Well, a conforming compiler is allowed to accept any code even if it's
ill-formed (see clause 1, subclause 1.4, for example).

6.7/3 allows to jump over a declaration of a local object if it is of POD
type and doesn't contain an initialiser. The code above is ill-formed.
A diagnostic is required; I conclude that based on the fact that the text
of the Standard does not contain "no diagnostic required" in 6.7/3.

V

 

[Victor Bazarov]

I will like to make my question more straight forward.

There is no question anywhere in this post. Care to ask it?

See code as
#include<iostream.h>

You've been told already, IIRC that <iostream.h> is a non-standard header.
Please stop using it, it's for your own good. Besides, in this program
there is no need to include it at all.

int main()
{
int a=1;
switch(a){
case 1:
int b=3;
break;
case 2:
break;

}
return 0;
}

When compiling, i get error:
"1.cpp", line 9: Error: This case bypasses initialization of a local
variable.

Yes, according to the rules of the language. There are no requirements
for a compiler to be as sophisticated as to prove that 'a' in your code
above cannot have any other value than 1. You can do it yourself, if you
need to. Some compilers can do it. Many don't bother.

The bottomline is, don't write such programs. And if the compiler tells
you that your code contains an error, fix it.

V