continue with switch

[v4vijayakumar]

'continue' within switch actually associated with the outer 'while'
loop. Is this behavior protable?

int ch = '\n';
while (true) {
switch(ch) {
case '\n': cout << "test"; continue;
}
}

the above loop executed endlessly in "gcc version 2.96 20000731 (Red
Hat Linux 7.3 2.96-110)".

[Richard Bos]

"v4vijayaku mar" <v4***********@ yahoo.com> wrote:

'continue' within switch actually associated with the outer 'while'
loop. Is this behavior protable?

int ch = '\n';
while (true) {
switch(ch) {
case '\n': cout << "test"; continue;
}
}

I have no idea if it's portable under C++, which is what your code
actually is. In this newsgroup we discuss ISO C, and yes, in that
language any continue statements do not apply to a switch statement, but
to the continue's closest surrounding loop statement, if there is any.

Richard

[Bart Rider]

v4vijayakumar wrote:

'continue' within switch actually associated with the outer 'while'
loop. Is this behavior protable?

int ch = '\n';
while (true) {
switch(ch) {
case '\n': cout << "test"; continue;
}
}

the above loop executed endlessly in "gcc version 2.96 20000731 (Red
Hat Linux 7.3 2.96-110)".

try this peace of code:
int main() {
int ch = 'a';
int i=0;
while (1) {
switch(ch) {
case 'a': printf("i=%d, a\n",i); ch='b'; continue; break;
case 'b': printf("i=%d, b\n",i); ch='c'; continue;
}
i++;

}
return 0;
}

you will see, that both printf-statements are executed. The continue
in switch is therefore to jump to the next case block, which is
executed *with* renewed comparison. The latter means, that if you
replace ch='b' by ch='d' or something, then only the first printf
is executed.

[Bart Rider]

Richard Bos wrote:

"v4vijayaku mar" <v4***********@ yahoo.com> wrote:

'continue' within switch actually associated with the outer 'while'
loop. Is this behavior protable?

int ch = '\n';
while (true) {
switch(ch) {
case '\n': cout << "test"; continue;
}
}

I have no idea if it's portable under C++, which is what your code
actually is. In this newsgroup we discuss ISO C, and yes, in that
language any continue statements do not apply to a switch statement, but
to the continue's closest surrounding loop statement, if there is any.

Richard

really?
Any break inside a switch statement is actually assigned to the
switch statement (it leaves the choice statement switch and continues
with the first statement after switch) and not for any surrounding loop.
The same applies to continue IMHO.

[Richard Heathfield]

Bart Rider said:

I have no idea if it's portable under C++, which is what your code
actually is. In this newsgroup we discuss ISO C, and yes, in that
language any continue statements do not apply to a switch statement, but
to the continue's closest surrounding loop statement, if there is any.

Richard
really?

Really.
Any break inside a switch statement is actually assigned to the
switch statement (it leaves the choice statement switch and continues
with the first statement after switch) and not for any surrounding loop.
The same applies to continue IMHO.

Really? Perhaps you'd better tell the ISO C Committee. I don't think they
took your opinion into account when signing off the C Standard. If you run,
you might still catch them.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)

[pete]

Bart Rider wrote:

v4vijayakumar wrote:

'continue' within switch actually associated with the outer 'while'
loop. Is this behavior protable?
while (true) {
switch(ch) { continue; }
}

You've got a continue statement inside a while loop.
It's just as simple as that.

try this peace of code:
int main() {

int ch = 'a';
int i=0;
while (1) {
switch(ch) {
case 'a': printf("i=%d, a\n",i); ch='b'; continue; break;
case 'b': printf("i=%d, b\n",i); ch='c'; continue;
}
i++;

}
return 0;
}

you will see, that both printf-statements are executed. The continue
in switch is therefore to jump to the next case block, which is
executed *with* renewed comparison.

That's not what happens at all.
You put the ++i loop counter in the wrong place
and bypassed it with the continue statement.

i=1, a
i=2, b


/* BEGIN new.c */

#include <stdio.h>

int main(void)
{
int ch = 'a';
int i = 0;

while (1) {
++i;
switch(ch) {
case 'a': printf("i=%d, a\n",i); ch='b'; continue; break;
case 'b': printf("i=%d, b\n",i); ch='c'; continue;
default: return 0;
}
}
return 0;
}

/* END new.c */
--
pete

[Flash Gordon]

Bart Rider wrote:

v4vijayakumar wrote:

'continue' within switch actually associated with the outer 'while'
loop. Is this behavior protable?

int ch = '\n';
while (true) {
switch(ch) {
case '\n': cout << "test"; continue;
}
}

the above loop executed endlessly in "gcc version 2.96 20000731 (Red
Hat Linux 7.3 2.96-110)".

try this peace of code:
int main() {

Better to be explicit about the lack of parameters:
int main(void)
int ch = 'a';
int i=0;
while (1) {
switch(ch) {
case 'a': printf("i=%d, a\n",i); ch='b'; continue; break;
The break above is misleading and/or useless because it will never be
reached.
case 'b': printf("i=%d, b\n",i); ch='c'; continue;
}
i++;

}
return 0;
}

you will see, that both printf-statements are executed. The continue
in switch is therefore to jump to the next case block, which is
executed *with* renewed comparison.
This is misleading at best. The continue is a jump to the end of the
loop (not the beginning, important to not for do ... while loops).
Obviously it then goes back to the start of the loop in the above code
and executes the switch again. Where the continue goes to has absolutely
*nothing* to do with case blocks.
The latter means, that if you
replace ch='b' by ch='d' or something, then only the first printf
is executed.

Your conclusion about your code is true. Add a printf before the switch
to see the rest is at best badly worded and in my opinion just plain wrong.

The description of continue in section 6.8.6.2 of the latest version of
the standard is very readable. You can find the latest draft at
http://www.open-std.org/jtc1/sc22/wg...docs/n1124.pdf
--
Flash Gordon, living in interesting times.
Web site - http://home.flash-gordon.me.uk/
comp.lang.c posting guidelines and intro:
http://clc-wiki.net/wiki/Intro_to_clc

[v4vijayakumar]

break/continue inside loop means
break-the-loop/continue-next-iteration, but inside switch (as it is not
associated with any looping) both could mean same thing, that is
execute-the-statement-next-to-switch.

am directly posting from google groups. can you please let me know,
what tool people generally use to read/post to groups. TIA.

[pete]

v4vijayakumar wrote:

break/continue inside loop means
break-the-loop/continue-next-iteration,
but inside switch (as it is not associated with any looping)
both could mean same thing, that is
execute-the-statement-next-to-switch.

No.

"continue" has no meaning related to "switch"

--
pete

[Richard Heathfield]

v4vijayakumar said:

break/continue inside loop means
break-the-loop/continue-next-iteration, but inside switch (as it is not
associated with any looping) both could mean same thing, that is
execute-the-statement-next-to-switch.
No, that is incorrect. A break will break out of the immediately enclosing
switch or loop, it's true - but continue has nothing to do with switch. It
is purely a loop construct.
am directly posting from google groups.
Actually, you are /indirectly/ posting from Google Groups. What happens is
that you compose your message, and post it to a Google Web server using
HTTP on port 80. Google then automatically transmits that article to a news
server using NNTP on port 119. So it's actually slightly indirect.

can you please let me know,
what tool people generally use to read/post to groups. TIA.

I use KNode. Some people use tin, trn, Free Agent, Netscape Messenger - in
fact, any newsreader you like.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)