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#include<stdio.h>
int main(void)
{
int a=1;
int b=2;
int c=0;
c=(a+b)++;
printf("%d",c);
return 0;
}

For the above program I got the error as Lvalue required.I compiled in
Turbo C/C++ compiler.

Can anyone please explain why the error is and how to eliminate it .

Thanks a lot.

Regards,
Raghu

raghu wrote:
> #include<stdio.h>
> int main(void)
> {
> int a=1;
> int b=2;
> int c=0;
> c=(a+b)++;
White space isn't expensive and makes code easier to read.

(a+b) is an expression, not a modifiable value. Why not c = a+b+1?

--
Ian Collins.

raghu said:

> #include<stdio.h>
> int main(void)
> {
> int a=1;
> int b=2;
> int c=0;
> c=(a+b)++;
> printf("%d",c);
> return 0;
> }
>
> For the above program I got the error as Lvalue required.

That's right, yes. ++ affects the value of an object. Arbitrary expressions
won't do.

> Can anyone please explain why the error is and how to eliminate it .

++ increases the value of an object by 1. (a+b) is not an object. To
eliminate the error, remove the ++:

#include<stdio.h>

int main(void)
{
int a = 1;
int b = 2;
int c = 0;

c = a + b;
printf("%d", c);

return 0;
}

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: normal service will be restored as soon as possible. Please do not
adjust your email clients.

Ian Collins said:

> raghu wrote:
>> #include<stdio.h>
>> int main(void)
>> {
>> int a=1;
>> int b=2;
>> int c=0;
>> c=(a+b)++;
> White space isn't expensive and makes code easier to read.
>
> (a+b) is an expression, not a modifiable value. Why not c = a+b+1?

Why the +1?

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: normal service will be restored as soon as possible. Please do not
adjust your email clients.

Richard Heathfield wrote:
> Ian Collins said:
>
>
>>raghu wrote:
>>
>>>#include<stdio.h>
>>>int main(void)
>>>{
>>>int a=1;
>>>int b=2;
>>>int c=0;
>>>c=(a+b)++;
>>
>>White space isn't expensive and makes code easier to read.
>>
>>(a+b) is an expression, not a modifiable value. Why not c = a+b+1?
>
> Why the +1?
>
I assumed he wanted to increment the sum of a and b.

--
Ian Collins.

Ian Collins said:

> Richard Heathfield wrote:
>> Ian Collins said:
>>
>>
>>>raghu wrote:
>>>
>>>>#include<stdio.h>
>>>>int main(void)
>>>>{
>>>>int a=1;
>>>>int b=2;
>>>>int c=0;
>>>>c=(a+b)++;
>>>
>>>White space isn't expensive and makes code easier to read.
>>>
>>>(a+b) is an expression, not a modifiable value. Why not c = a+b+1?
>>
>> Why the +1?
>>
> I assumed he wanted to increment the sum of a and b.

How would that affect the value of c?

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: normal service will be restored as soon as possible. Please do not
adjust your email clients.

Richard Heathfield wrote:
> Ian Collins said:
>
>
>>Richard Heathfield wrote:
>>
>>>Ian Collins said:
>>>
>>>
>>>
>>>>raghu wrote:
>>>>
>>>>
>>>>>#include<stdio.h>
>>>>>int main(void)
>>>>>{
>>>>>int a=1;
>>>>>int b=2;
>>>>>int c=0;
>>>>>c=(a+b)++;
>>>>
>>>>White space isn't expensive and makes code easier to read.
>>>>
>>>>(a+b) is an expression, not a modifiable value. Why not c = a+b+1?
>>>
>>>Why the +1?
>>>
>>
>>I assumed he wanted to increment the sum of a and b.
>
>
> How would that affect the value of c?
>
I assumed he wanted to increment the sum of a and b and assign it to c.

--
Ian Collins.

Ian Collins said:

> Richard Heathfield wrote:
>>
> I assumed he wanted to increment the sum of a and b and assign it to c.

If so, would he not have written ++(a + b) ?

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: normal service will be restored as soon as possible. Please do not
adjust your email clients.

Richard Heathfield wrote:
> Ian Collins said:
>
>
>>Richard Heathfield wrote:
>>
>>I assumed he wanted to increment the sum of a and b and assign it to c.
>
>
> If so, would he not have written ++(a + b) ?
>
Why?

--
Ian Collins.

Ian Collins wrote:
> Richard Heathfield wrote:
> > Ian Collins said:
> >
> >
> >>Richard Heathfield wrote:
> >>
> >>I assumed he wanted to increment the sum of a and b and assign it to c.
> >
> >
> > If so, would he not have written ++(a + b) ?
> >
> Why?

I think Richard's point was that the postfix "++" implies adding one
_after_ c had been assigned the value of "a+b"... Quite what the one
would be added to remains a mystery.