• C (22/22) Feb 02 2004 Should this be allowed ?
• Manfred Nowak (12/17) Feb 02 2004 [...]
• C (18/35) Feb 02 2004 Hmm ? I was refering to arbitrary code within the switch statement, not
• davepermen (5/45) Feb 03 2004 switch is just a shortcut for a lot of goto's to the different case-labe...
• J Anderson (18/23) Feb 03 2004 I didn't realize that. This means no-fall-through cases are possible
• Manfred Nowak (8/9) Feb 05 2004 It is correct code. If you look at the docs then you will see, that

"C" <dont respond.com> writes:

Should this be allowed ?
--

import std.c.stdio;

void main ()
{

 const int option = 3;
 int assignAnInt = 0;

 switch (  option )
 {

  if (  true )
  {
   puts( "never executed" ) ;
  }

  assignAnInt = 21;

  case 3:
  puts( "3" );
  break;
  default:
  break;


 }

 printf( "%d",assignAnInt );

}

Manfred Nowak <svv1999 hotmail.com> writes:

C wrote:

 Should this be allowed ?

[...]
  switch (  option )
  {
   assignAnInt = 21;
   case 3:

[...]

Yes.

It is the same as with:

void main(){
 if( true )
  printf("Not executed.\n");
 else
  printf("Executed.\n");
}

So long.

"C" <dont respond.com> writes:

Hmm ?  I was refering to arbitrary code within the switch statement, not
belonging to a case .  Its not correct code I thought the compiler would
complain , however that also seems to be legal in C++.

#include <cstdio>
#include <cstdlib>

int main () {
 int option = 4;

 switch ( option ) {
  if ( option == 4 ) exit(0);

 case 4: puts("here");
 }

 return 1;
}

outputs : here

shrug,
C
"Manfred Nowak" <svv1999 hotmail.com> wrote in message
news:bvlf1b$m16$1 digitaldaemon.com...
 C wrote:

 Should this be allowed ?

 [...]
  switch (  option )
  {
   assignAnInt = 21;
   case 3:

 [...]

 Yes.

 It is the same as with:

 void main(){
  if( true )
   printf("Not executed.\n");
  else
   printf("Executed.\n");
 }

 So long.

"davepermen" <davepermen hotmail.com> writes:

switch is just a shortcut for a lot of goto's to the different case-labels.
so the behaviour is entierly understandable. it jumps over the rest to the
first case that fits..

"C" <dont respond.com> schrieb im Newsbeitrag
news:bvmvfq$58e$1 digitaldaemon.com...
 Hmm ?  I was refering to arbitrary code within the switch statement, not
 belonging to a case .  Its not correct code I thought the compiler would
 complain , however that also seems to be legal in C++.

 #include <cstdio>
 #include <cstdlib>

 int main () {
  int option = 4;

  switch ( option ) {
   if ( option == 4 ) exit(0);

  case 4: puts("here");
  }

  return 1;
 }

 outputs : here

 shrug,
 C
 "Manfred Nowak" <svv1999 hotmail.com> wrote in message
 news:bvlf1b$m16$1 digitaldaemon.com...
 C wrote:

 Should this be allowed ?

 [...]
  switch (  option )
  {
   assignAnInt = 21;
   case 3:

 [...]

 Yes.

 It is the same as with:

 void main(){
  if( true )
   printf("Not executed.\n");
  else
   printf("Executed.\n");
 }

 So long.

J Anderson <REMOVEanderson badmama.com.au> writes:

C wrote:

Hmm ?  I was refering to arbitrary code within the switch statement, not
belonging to a case .  Its not correct code I thought the compiler would
complain , however that also seems to be legal in C++.
<snip>
  

I didn't  realize that.  This means no-fall-through cases are possible 
in C++ (and C) ie

#define when(T) break;case T:

int main(int argc, char* argv[])
{
    int X = 5;
    switch (5)
    {
    when(5)
        printf("5\n");
    when(1)
        printf("1\n");
    }
    return 0;
}

-- 
-Anderson: http://badmama.com.au/~anderson/

Manfred Nowak <svv1999 hotmail.com> writes:

C wrote:

 Its not correct code

It is correct code. If you look at the docs then you will see, that
switchstatement derives to blockstatement and not something like
casesequence.

So consecutive breakstatements are also correct. Also code between two
breakstatements without a casestatement. And code after the last
breakstatement also.

So long.