a basic question,need help

[jackie]

i know that in c plus plus,++x returns l-value while x++ returns r-
value,but what is the situation in c,are both ++x and x++ return r-
value? i don't know how C99 defines it,thx.

[Eric Sosman]

Re: a basic question,need help

jackie wrote:

i know that in c plus plus,++x returns l-value while x++ returns r-
value,but what is the situation in c,are both ++x and x++ return r-
value? i don't know how C99 defines it,thx.

In C (all versions), neither ++x nor x++ is an lvalue.
I'm surprised to hear that this is different in C++, because
it makes no sense to me. Is

int x = 42;
++x = 97;

legal in C++? If so, what is the value of x afterwards?
(Actually, you can disregard the second question: If the
answer to the first is anything other than "No," I don't
want to know anything more about C++.)

--
Eric Sosman
esosman@ieee-dot-org.invalid

[Tommy]

Re: a basic question,need help

[cuitao@mytest c]$ vi pp.c

#include <stdio.h>
int main()
{
int x=10;
x=20;
printf("\nx=%d\ n",x);

return 0;
}
~
~
~
"pp.c" 9L, 99C written
[cuitao@mytest c]$
[cuitao@mytest c]$
[cuitao@mytest c]$ cc ./pp.c -o ./pp
[cuitao@mytest c]$
[cuitao@mytest c]$ ./pp

x=20
[cuitao@mytest c]$


[cuitao@mytest c]$ vi pp.c

#include <stdio.h>
int main()
{
int x=10;
x++=20;
printf("\nx=%d\ n",x);

return 0;
}
~
~
~
"pp.c" 9L, 101C written
[cuitao@mytest c]$
[cuitao@mytest c]$
[cuitao@mytest c]$ cc ./pp.c -o ./pp
../pp.c: In function `main':
../pp.c:5: error: invalid lvalue in assignment





[cuitao@mytest c]$ vi pp.c

#include <stdio.h>
int main()
{
int x=10;
++x=20;
printf("\nx=%d\ n",x);

return 0;
}
~
~
~
"pp.c" 9L, 101C written
[cuitao@mytest c]$ cc ./pp.c -o ./pp
../pp.c: In function `main':
../pp.c:5: error: invalid lvalue in assignment
[cuitao@mytest c]$







"Eric Sosman" <esosman@ieee-dot-org.invalid>
??????:4sadnefS S4WSyT_VnZ2dnUV Z_jOdnZ2d@comca st.com...

jackie wrote:

>i know that in c plus plus,++x returns l-value while x++ returns r-
>value,but what is the situation in c,are both ++x and x++ return r-
>value? i don't know how C99 defines it,thx.

>
In C (all versions), neither ++x nor x++ is an lvalue.
I'm surprised to hear that this is different in C++, because
it makes no sense to me. Is
>
int x = 42;
++x = 97;
>
legal in C++? If so, what is the value of x afterwards?
(Actually, you can disregard the second question: If the
answer to the first is anything other than "No," I don't
want to know anything more about C++.)
>
--
Eric Sosman
esosman@ieee-dot-org.invalid

[Ian Collins]

Re: a basic question,need help

Eric Sosman wrote:

jackie wrote:

>i know that in c plus plus,++x returns l-value while x++ returns r-
>value,but what is the situation in c,are both ++x and x++ return r-
>value? i don't know how C99 defines it,thx.

>
In C (all versions), neither ++x nor x++ is an lvalue.
I'm surprised to hear that this is different in C++, because
it makes no sense to me. Is
>
int x = 42;
++x = 97;
>
legal in C++?

No.

--
Ian Collins.

[CBFalconer]

Re: a basic question,need help

jackie wrote:

>
i know that in c plus plus,++x returns l-value while x++ returns
r- value,but what is the situation in c,are both ++x and x++
return r- value? i don't know how C99 defines it,thx.

You're wrong. Illustrative snippet:

volatile int x = 123;
int y;

y = x;
if ((y == x++) && ((y + 1) == x)) puts("Correct x++");
else puts("Bad system");

y = x;
if (((y + 1) == ++x) && ((y + 1) == x)) puts("Correct ++x");
else puts("Bad system");

The operator has nothing to to with lvalues and rvalues. Same in
C++. Note that in the snippet the first phrase in the test (before
the &&) is completed before the 2nd phase is begun.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.

[Daniel Pitts]

Re: a basic question,need help

Ian Collins wrote:

Eric Sosman wrote:

>jackie wrote:

>>i know that in c plus plus,++x returns l-value while x++ returns r-
>>value,but what is the situation in c,are both ++x and x++ return r-
>>value? i don't know how C99 defines it,thx.

> In C (all versions), neither ++x nor x++ is an lvalue.
>I'm surprised to hear that this is different in C++, because
>it makes no sense to me. Is
>>
> int x = 42;
> ++x = 97;
>>
>legal in C++?

>
No.
>

Sure it is:

class foo {
public:
foo(int x) {
}
foo &operator=(i nt x){}
};
foo &operator++( foo &x) {
return x;
}

int main() {
#define int foo
int x = 72;
++x = 32;
}

--
Daniel Pitts' Tech Blog: <http://virtualinfinity .net/wordpress/>

[Eric Schmidt]

Re: a basic question,need help

Eric Sosman wrote:

jackie wrote:
>

>i know that in c plus plus,++x returns l-value while x++ returns r-
>value,but what is the situation in c,are both ++x and x++ return r-
>value? i don't know how C99 defines it,thx.

>
>
In C (all versions), neither ++x nor x++ is an lvalue.
I'm surprised to hear that this is different in C++, because
it makes no sense to me. Is
>
int x = 42;
++x = 97;
>
legal in C++? If so, what is the value of x afterwards?
(Actually, you can disregard the second question: If the
answer to the first is anything other than "No," I don't
want to know anything more about C++.)
>

It yields undefined behavior in C++ because x is modified twice without
an intervening sequence point. But something like &(++x) is legal
(though it seems useless to me).
** Posted from http://www.teranews.com **

[Keith Thompson]

Re: a basic question,need help

CBFalconer <cbfalconer@yah oo.comwrites:

jackie wrote:

>>
>i know that in c plus plus,++x returns l-value while x++ returns
>r- value,but what is the situation in c,are both ++x and x++
>return r- value? i don't know how C99 defines it,thx.

>
You're wrong.

Just what is it that you think he's wrong about?

(I'm assuming jackie is a "he"; apologies if I'm mistaken.)

In C++, the result of ``++x'' is an lvalue, and the result of ``x++''
is not an lvalue, so he's right about that; it's off-topic, but a
perfectly reasonable introduction to his question.

The rest of the post was a question, which can hardly be wrong, and a
statement that he doesn't know something, which you can hardly assume
is wrong.

Illustrative snippet:

Illustrative of what?

volatile int x = 123;
int y;
>
y = x;
if ((y == x++) && ((y + 1) == x)) puts("Correct x++");
else puts("Bad system");
>
y = x;
if (((y + 1) == ++x) && ((y + 1) == x)) puts("Correct ++x");
else puts("Bad system");

What is the point of declaring x volatile?

The operator has nothing to to with lvalues and rvalues. Same in
C++. Note that in the snippet the first phrase in the test (before
the &&) is completed before the 2nd phase is begun.

Certainly it does.

In both C and C++, the operand of a prefix or postfix "++" must be a
modifiable lvalue. In C, the result of either operator is not an
lvalue; in C++, one is an lvalue and one isn't.

In any case, your code snippet is equally valid in either C or C++,
and would behave in exactly the same way even if both prefix and
postfix "++" yielded lvalues; you never use them in a context that
requires an lvalue.

So what was your point again?

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[jackie]

Re: a basic question,need help

On Aug 13, 3:29 pm, Keith Thompson <ks...@mib.orgw rote:

CBFalconer <cbfalco...@yah oo.comwrites:

jackie wrote:

>

i know that in c plus plus,++x returns l-value while x++ returns
r- value,but what is the situation in c,are both ++x and x++
return r- value? i don't know how C99 defines it,thx.

>

You're wrong.

>
Just what is it that you think he's wrong about?
>
(I'm assuming jackie is a "he"; apologies if I'm mistaken.)
>
In C++, the result of ``++x'' is an lvalue, and the result of ``x++''
is not an lvalue, so he's right about that; it's off-topic, but a
perfectly reasonable introduction to his question.
>
The rest of the post was a question, which can hardly be wrong, and a
statement that he doesn't know something, which you can hardly assume
is wrong.
>

               Illustrative snippet:

>
Illustrative of what?
>

    volatile int x = 123;
    int          y;

>

    y = x;
    if ((y == x++) && ((y + 1) == x)) puts("Correct x++");
    else                              puts("Bad system");

>

    y = x;
    if (((y + 1) == ++x) && ((y + 1) == x)) puts("Correct ++x");
    else                                    puts("Bad system");

>
What is the point of declaring x volatile?
>

The operator has nothing to to with lvalues and rvalues.  Same in
C++.  Note that in the snippet the first phrase in the test (before
the &&) is completed before the 2nd phase is begun.

>
Certainly it does.
>
In both C and C++, the operand of a prefix or postfix "++" must be a
modifiable lvalue.  In C, the result of either operator is not an
lvalue; in C++, one is an lvalue and one isn't.
>
In any case, your code snippet is equally valid in either C or C++,
and would behave in exactly the same way even if both prefix and
postfix "++" yielded lvalues; you never use them in a context that
requires an lvalue.
>
So what was your point again?
>
--
Keith Thompson (The_Other_Keit h) ks...@mib.org  <http://www.ghoti.net/~kst>
Nokia
"We must do something.  This is something.  Therefore, we must do this."
    -- Antony Jay and Jonathan Lynn, "Yes Minister"

u're right,i'm a "he". ^-^
and i think u have already give me ur answer,that's both ++x and x++
return r-value in c,do u mean that? thx for ur help

[CBFalconer]

Re: a basic question,need help

Keith Thompson wrote:

CBFalconer <cbfalconer@yah oo.comwrites:
>

.... snip ...

>

>Illustrative snippet:

>
Illustrative of what?

Of the behaviour of x++ and ++x.

>

> volatile int x = 123;
> int y;
>>
> y = x;
> if ((y == x++) && ((y + 1) == x)) puts("Correct x++");
> else puts("Bad system");
>>
> y = x;
> if (((y + 1) == ++x) && ((y + 1) == x)) puts("Correct ++x");
> else puts("Bad system");

>
What is the point of declaring x volatile?

That ensures that x is accessed for each equality test. That way
it can't be carried along in a working register.

>

>The operator has nothing to do with lvalues and rvalues. Same in
>C++. Note that in the snippet the first phrase in the test (before
>the &&) is completed before the 2nd phase is begun.

>
Certainly it does.
>
In both C and C++, the operand of a prefix or postfix "++" must be a
modifiable lvalue. In C, the result of either operator is not an
lvalue; in C++, one is an lvalue and one isn't.
>
In any case, your code snippet is equally valid in either C or C++,
and would behave in exactly the same way even if both prefix and
postfix "++" yielded lvalues; you never use them in a context that
requires an lvalue.
>
So what was your point again?

To demonstrate the actions of x++ and ++x, and their difference.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.

[James Kuyper]

Re: a basic question,need help

CBFalconer wrote:

Keith Thompson wrote:

>CBFalconer <cbfalconer@yah oo.comwrites:
>>

... snip ...

>>Illustrativ e snippet:

>Illustrative of what?

>
Of the behaviour of x++ and ++x.

>> volatile int x = 123;
>> int y;
>>>
>> y = x;
>> if ((y == x++) && ((y + 1) == x)) puts("Correct x++");
>> else puts("Bad system");
>>>
>> y = x;
>> if (((y + 1) == ++x) && ((y + 1) == x)) puts("Correct ++x");
>> else puts("Bad system");

....

>In any case, your code snippet is equally valid in either C or C++,
>and would behave in exactly the same way even if both prefix and
>postfix "++" yielded lvalues; you never use them in a context that
>requires an lvalue.
>>
>So what was your point again?

>
To demonstrate the actions of x++ and ++x, and their difference.

The question was about the lvalue-ness of x++ and ++x; what does your
example have to do with that question?

[CBFalconer]

Re: a basic question,need help

James Kuyper wrote:

CBFalconer wrote:

>Keith Thompson wrote:

>>CBFalconer <cbfalconer@yah oo.comwrites:
>>>

>... snip ...
>>

>>>Illustrati ve snippet:
>>Illustrativ e of what?

>>
>Of the behaviour of x++ and ++x.

>>>
>>> volatile int x = 123;
>>> int y;
>>>>
>>> y = x;
>>> if ((y == x++) && ((y + 1) == x)) puts("Correct x++");
>>> else puts("Bad system");
>>>>
>>> y = x;
>>> if (((y + 1) == ++x) && ((y + 1) == x)) puts("Correct ++x");
>>> else puts("Bad system");

...

>>In any case, your code snippet is equally valid in either C or
>>C++, and would behave in exactly the same way even if both
>>prefix and postfix "++" yielded lvalues; you never use them in
>>a context that requires an lvalue.
>>>
>>So what was your point again?

>>
>To demonstrate the actions of x++ and ++x, and their difference.

>
The question was about the lvalue-ness of x++ and ++x; what does
your example have to do with that question?

Well, for one thing you can see immediately that x++ can't have an
lvalue. There is nothing holding that value, apart from the
reference y.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.

[christian.bau]

Re: a basic question,need help

On Aug 13, 4:25 am, Eric Sosman <esos...@ieee-dot-org.invalidwrot e:

I'm surprised to hear that this is different in C++, because
it makes no sense to me. Is
>
int x = 42;
++x = 97;
>
legal in C++? If so, what is the value of x afterwards?
(Actually, you can disregard the second question: If the
answer to the first is anything other than "No," I don't
want to know anything more about C++.)

It is syntactically legal, but it invokes undefined behaviour same as
it would in C if ++x was an lvalue (the same object is modified twice
without intervening sequence point). Something that is legal and makes
a bit of sense would be:

int x = 42;
int *p = &++x;

This will give you x = 43, and p = &x. There is motivation for this
because of "references " in C++. If you have a C++ function

int f (int& x) { return x++; }

then you can call

int x = 42;
int y = f (++x);

which will increase x by one, pass x "by reference" to the function f
which is rather the same as passing the address; f will increase x in
the caller (because it has a reference to it) and return the value
before the incremenent. Effectively this will set x to 44 and y to 43,
and there is no undefined behavior because there are plenty of
sequence points.

[Keith Thompson]

Re: a basic question,need help

jackie <jackiexxm@gmai l.comwrites:
[...]

u're right,i'm a "he". ^-^
and i think u have already give me ur answer,that's both ++x and x++
return r-value in c,do u mean that? thx for ur help

Yes, both ++x and x++ yield a non-lvalue in C. (You can call that an
"rvalue", but C doesn't use the term except in a single footnote.)

Incidentally, silly abbreviations like "u", "ur", and "thx" are
frowwned upon here. Please take the time to spell out simple words
like "you", "your", and "thanks". Not everyone here speaks English as
a native language -- and those of us who do generally prefer to read
actual English.

Also, when you post a followup, it's rarely necessary to quote the
entire parent article. Delete any quoted material that's not relevant
to your followup, leaving just enough so that your article makes sense
to someone who hasn't seen the parent. In particular, don't quote
signatures unless you're actually commenting on them.

Thx. 8-)}

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[jameskuyper@verizon.net]

Re: a basic question,need help

CBFalconer wrote:

James Kuyper wrote:

CBFalconer wrote:

Keith Thompson wrote:
>CBFalconer <cbfalconer@yah oo.comwrites:
>>
... snip ...
>
>>Illustrativ e snippet:
>Illustrative of what?
>
Of the behaviour of x++ and ++x.
>>
>> volatile int x = 123;
>> int y;
>>>
>> y = x;
>> if ((y == x++) && ((y + 1) == x)) puts("Correct x++");
>> else puts("Bad system");
>>>
>> y = x;
>> if (((y + 1) == ++x) && ((y + 1) == x)) puts("Correct ++x");
>> else puts("Bad system");

...

>In any case, your code snippet is equally valid in either C or
>C++, and would behave in exactly the same way even if both
>prefix and postfix "++" yielded lvalues; you never use them in
>a context that requires an lvalue.
>>
>So what was your point again?
>
To demonstrate the actions of x++ and ++x, and their difference.

The question was about the lvalue-ness of x++ and ++x; what does
your example have to do with that question?

>
Well, for one thing you can see immediately that x++ can't have an
lvalue. There is nothing holding that value, apart from the
reference y.

Actually, it is ++x, not x++, which is an lvalue in C++, and your code
does nothing to demonstrate that "There is nothing holding" the value
of ++x, because you code makes no use of ++x in a context that
requires an lvalue.
The example provided by Daniel Pits does use ++x in a context that
requires an lvalue. The fact that his code compiles and works as he
expected it to work demonstrates that jackie was in fact correct in
saying that ++x is an lvalue in C++, your assertions to the contrary
notwithstanding .

Actually, as a demonstration, Daniel's code could be improved upon. It
would be clearer what the lvalued-ness of ++x means in C++, if his foo
class had contained a non-static data member that participated in
operator++() and operator=() in a non-trivial fashion. I assure you
that adding such a data member would not prevent the code from
working. If he had done so, it would have been quite clear that there
is indeed something "holding that value". The "something" is what C++
calls an temporary object.

The fact that C cannot usefully make ++x an lvalue is due the fact
that C structs cannot have member functions. The fact that C has no
need to make ++x an lvalue is mainly due to the fact that C doesn't
have operator overloads. It's not because it would be impossible to
make ++x an lvalue in C; it would have been just as feasible for C to
introduce the concept of a temporary object.