?: as an lvalue

[Rahul]

Hi Everyone,

I have the following piece of code, and i expected an error, however
i don't get an error,

int main()
{
int aa=0,b=0;
1>0?aa:b = 10;
printf("value %d %d\n",aa,b);
return(0);
}

and output is value 0 0

Thanks in advance ! ! !

[Kenny McCormack]

Re: ?: as an lvalue

In article <f840856d-216c-4bab-ac4a-f962f88a2c3a@h1 1g2000prf.googl egroups.com>,
Rahul <sam_cit@yahoo. co.inwrote:

>Hi Everyone,
>
I have the following piece of code, and i expected an error, however
>i don't get an error,
>
>int main()
>{
> int aa=0,b=0;
> 1>0?aa:b = 10;
printf("value %d %d\n",aa,b);
return(0);
>}
>
>and output is value 0 0
>
>Thanks in advance ! ! !

int main(void) is better
failure to include stdio.h
unnecessarily parenthesizing the return value
failure to indent the return statement properly
meaningless variable names
failure to capitalize "i" in text (e.g., "i don't get an error")
excessive use of exclamation points
improper capitalization of word "everyone"
use of tacky expression "Thanks in advance"

[Richard Heathfield]

Re: ?: as an lvalue

Rahul said:

Hi Everyone,
>
I have the following piece of code, and i expected an error, however
i don't get an error,
>
int main()
{
int aa=0,b=0;
1>0?aa:b = 10;

This is a syntax error (which therefore requires the implementation to
diagnose it as such). The syntax of the conditional operator is:

conditional-expression:
logical-OR-expression
logical-OR-expression ? expression : conditional-expression

which your code violates.

printf("value %d %d\n",aa,b);
return(0);
}
>
and output is value 0 0
>
Thanks in advance ! ! !

Turn up your warning level. Which implementation are you using?

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[Rahul]

Re: ?: as an lvalue

On Mar 30, 5:19 pm, Richard Heathfield <r...@see.sig.i nvalidwrote:

Rahul said:
>

Hi Everyone,

>

I have the following piece of code, and i expected an error, however
i don't get an error,

>

int main()
{
int aa=0,b=0;
1>0?aa:b = 10;

>
This is a syntax error (which therefore requires the implementation to
diagnose it as such). The syntax of the conditional operator is:
>
conditional-expression:
logical-OR-expression
logical-OR-expression ? expression : conditional-expression
>
which your code violates.
>

printf("value %d %d\n",aa,b);
return(0);
}

>

and output is value 0 0

>

Thanks in advance ! ! !

>
Turn up your warning level. Which implementation are you using?
>
--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

I tried MS VC++ 8.0...

http://www.dinkumware.com/exam/default.aspx

[Flash Gordon]

Re: ?: as an lvalue

Rahul wrote, On 30/03/08 12:05:

Hi Everyone,
>
I have the following piece of code, and i expected an error, however
i don't get an error,

In that case you need to tell your compiler to act as a C compiler
rather than as an extended-C-like-language compiler.

int main()
{
int aa=0,b=0;
1>0?aa:b = 10;

This is illegal and a C compiler is required to produce a diagnostic
(warning, error or whatever).

printf("value %d %d\n",aa,b);

Not related to your current problem, but you need to include stdio.h to
use printf.

return(0);
}
>
and output is value 0 0

If you want to do this then a legal way is as follows:
#include <stdio.h>

int main()
{
int aa=0,b=0;
*(1>0?&aa:&b) = 10;
printf("value %d %d\n",aa,b);
return(0);
}
--
Flash Gordon

[Richard Heathfield]

Re: ?: as an lvalue

Rahul said:

On Mar 30, 5:19 pm, Richard Heathfield <r...@see.sig.i nvalidwrote:

>Rahul said:
>>

Hi Everyone,

>>

I have the following piece of code, and i expected an error, however
i don't get an error,

>>

int main()
{
int aa=0,b=0;
1>0?aa:b = 10;

>>
>This is a syntax error (which therefore requires the implementation to
>diagnose it as such). The syntax of the conditional operator is:
>>
> conditional-expression:
> logical-OR-expression
> logical-OR-expression ? expression : conditional-expression
>>
>which your code violates.
>>

printf("value %d %d\n",aa,b);
return(0);
}

>>

and output is value 0 0

>>

Thanks in advance ! ! !

>>
>Turn up your warning level. Which implementation are you using?

>
I tried MS VC++ 8.0...

I think the switches you'll need for Visual C are -W4 -Za

(-W4 turns the warning level up as far as it'll go - unless they've added
another level in the last few years - and -Za switches off Microsoft
extensions.)

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[Andrey Tarasevich]

Re: ?: as an lvalue

Rahul wrote:

I tried MS VC++ 8.0...

I tried the same compiler and got essentially the same error, saying that left
operand must be a lvalue.

http://www.dinkumware.com/exam/default.aspx

If you tried it at this link, then most likely your code was compiled as C++. In
C++ it would be parsed as '1 0 ? aa : (b = 10)'. Additionally in C++ the
result of ?: can be lvalue. It is not surprise that you didn't get an error.

--
Best regards,
Andrey Tarasevich

[Richard Heathfield]

Re: ?: as an lvalue

Flash Gordon said:

Rahul wrote, On 30/03/08 12:05:

>Hi Everyone,
>>
> I have the following piece of code, and i expected an error, however
>i don't get an error,

>
In that case you need to tell your compiler to act as a C compiler
rather than as an extended-C-like-language compiler.

The OP is using http://www.dinkumware.com/exam/default.aspx to test his
code. When I present that site with the following source:

#include <stdio.h>

int main(void)
{
int a = 0;
int b = 0;

1 0 ? a : b = 10;

printf("%d %d\n", a, b);

return 0;
}

and select the only C option I can find, which is the EDG C99 option, I get
no diagnostic messages whatsoever (unless you count "Code compiled
successfully!" as a diagnostic message).

Either this is a bug in EDG's compiler, or the rules changed in C99. I can
find no evidence of a rule change in C99.

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[Dave Hansen]

Re: ?: as an lvalue

On Mar 30, 7:19 am, Richard Heathfield <r...@see.sig.i nvalidwrote:

Rahul said:
>

Hi Everyone,

>

I have the following piece of code, and i expected an error, however
i don't get an error,

>

int main()
{
int aa=0,b=0;
1>0?aa:b = 10;

>
This is a syntax error (which therefore requires the implementation to
diagnose it as such). The syntax of the conditional operator is:
>

I believe the syntax is legal. It just doesn't do what the OP wanted
it to.

Consider that it parses as

(1>0)?(aa):(b=1 0);

Change it to 1<0?aa:b=10; and see the result...

Regards,

-=Dave

[Eric Sosman]

Re: ?: as an lvalue

Dave Hansen wrote:

On Mar 30, 7:19 am, Richard Heathfield <r...@see.sig.i nvalidwrote:

>Rahul said:
>>

>>Hi Everyone,
>> I have the following piece of code, and i expected an error, however
>>i don't get an error,
>>int main()
>>{
>>int aa=0,b=0;
>>1>0?aa:b = 10;

>This is a syntax error (which therefore requires the implementation to
>diagnose it as such). The syntax of the conditional operator is:
>>

>
I believe the syntax is legal. It just doesn't do what the OP wanted
it to.
>
Consider that it parses as
>
(1>0)?(aa):(b=1 0);

Would you care to place a small wager on that?

You could refer to the formal grammar in the Standard
to settle the question, or you could ask informally which
of = and ?: "binds more tightly." The parse you suggest
would follow if = binds more tightly, in which case

x = a ? b : c;

would parse as

(x = a) ? b : c;

Since we know that it actually parses as

x = (a ? b : c);

you may be about to lose some money ...

--
Eric.Sosman@sun .com

[Andrey Tarasevich]

Re: ?: as an lvalue

Richard wrote:

...
But (c?x:y)=v;
I dont really know what to say.

Is there any reason why you believe that the property of "being an
lvalue" should be necessarily lost in the process of selection from two
lvalues of the same type?

I mean I'm OK personally with the way it works in C. I just like to know
what is it exactly in '(c?x:y)=v' that triggers a "I don't really know
what to say" reaction from some people.

--
Best regards,
Andrey Tarasevich

[Richard]

Re: ?: as an lvalue

Andrey Tarasevich <andreytarasevi ch@hotmail.comw rites:

Richard wrote:

>...
>But (c?x:y)=v;
>I dont really know what to say.

>
Is there any reason why you believe that the property of "being an
lvalue" should be necessarily lost in the process of selection from
two lvalues of the same type?
>
I mean I'm OK personally with the way it works in C. I just like to
know what is it exactly in '(c?x:y)=v' that triggers a "I don't really
know what to say" reaction from some people.

because its not a macro? it returns a value. I dont know the legalise
words but it seems "obvious" enough to me, but again it might be because
I am tainted.

[Willem]

Re: ?: as an lvalue

Eric wrote:
) Dave Hansen wrote:
)Consider that it parses as
)>
) (1>0)?(aa):(b=1 0);
)
) Would you care to place a small wager on that?
)
) You could refer to the formal grammar in the Standard
) to settle the question, or you could ask informally which
) of = and ?: "binds more tightly." The parse you suggest
) would follow if = binds more tightly, in which case

Technically, it would also follow if = binds equally tightly.
Left-to-right and all that.

) x = a ? b : c;
)
) would parse as
)
) (x = a) ? b : c;

Again, purely technically, it would not if = were to bind equally tightly.

) Since we know that it actually parses as
)
) x = (a ? b : c);
)
) you may be about to lose some money ...

I'm not going to bet on it though.


SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT

[Willem]

Re: ?: as an lvalue

Andrey wrote:
) Richard wrote:
) ...
)But (c?x:y)=v;
)I dont really know what to say.
)
) Is there any reason why you believe that the property of "being an
) lvalue" should be necessarily lost in the process of selection from two
) lvalues of the same type?
)
) I mean I'm OK personally with the way it works in C. I just like to know
) what is it exactly in '(c?x:y)=v' that triggers a "I don't really know
) what to say" reaction from some people.

Well then why not also make it possible for functions (that return
pointers) to be lvalues ?

returnspointert ostruct(foo)->bar = baz;

Or is that already legal ?


SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT

[Andrey Tarasevich]

Re: ?: as an lvalue

Richard wrote:

>...
>I mean I'm OK personally with the way it works in C. I just like to
>know what is it exactly in '(c?x:y)=v' that triggers a "I don't really
>know what to say" reaction from some people.

>
because its not a macro? it returns a value. I dont know the legalise
words but it seems "obvious" enough to me, but again it might be because
I am tainted.

Well, unary '*' operator is also not a macro. Yet it evaluates to an
lvalue. Same for '[]' operator (by definition). Do you find this strange
as well?

--
Best regards,
Andrey Tarasevich