operator has no effect

[onkar]

#include<stdio. h>
int main(void){
unsigned int m=0xfdff;
~m;
printf("%x\n",m );
return 0;
}


gives me :
$ gcc -Wall -g -o test test.c
test.c: In function `main':
test.c:4: warning: statement with no effect
$ ./test
fdff


whereas this works out properly :

#include<stdio. h>
int main(void){
unsigned int m=0xfdff;
printf("%x\n",~ m);
return 0;
}

$gcc -Wall -g -o test test.c
$ ./test
ffff0200


Why ???

[sam_cit@yahoo.co.in]

Re: operator has no effect

#include<stdio. h>

int main(void){
unsigned int m=0xfdff;
~m;
printf("%x\n",m );
return 0;

Try this,

m=~m;

[onkar]

Re: operator has no effect

I mean than why did the second case work ??

sam_...@yahoo.c o.in wrote:

#include<stdio. h>
int main(void){
unsigned int m=0xfdff;
~m;
printf("%x\n",m );
return 0;

>
>
Try this,
>
m=~m;

[Ian Collins]

Re: operator has no effect

onkar wrote:

Please don't top post.

sam_...@yahoo.c o.in wrote:
>

>>>#include<std io.h>
>>>int main(void){
>> unsigned int m=0xfdff;
>> ~m;
>> printf("%x\n",m );
>> return 0;

>>
>>
> Try this,
>>
> m=~m;

>
I mean than why did the second case work ??
>

Because you were passing ~m to printf. ~m on its own doesn't change m,
it just evaluates the expression and discards the result.

--
Ian Collins.

[Zara]

Re: operator has no effect

On 13 Nov 2006 22:14:50 -0800, "onkar" <onkar.n.m@gmai l.comwrote:

>#include<stdio .h>
>int main(void){
unsigned int m=0xfdff;
~m;

You complement m and do nothing with result. As complementing an
unsigned int has no effects, then the operator has no effect.

printf("%x\n",m );
return 0;
>}
>
>
>gives me :

<...>

>test.c:4: warning: statement with no effect

<...>

>
>whereas this works out properly :
>
>#include<stdio .h>
>int main(void){
unsigned int m=0xfdff;
printf("%x\n",~ m);

You now complement m and use its result to pass it to printf, thus the
operator serves some clear purpose and does have effect.

return 0;
>}
>

Regrads,

Zara

[Richard Heathfield]

Re: operator has no effect

onkar said:

#include<stdio. h>
int main(void){
unsigned int m=0xfdff;
~m;

This is an expression, and C likes to evaluate expressions. In this case,
the expression ~m is evaluated, and you do nothing with the result. m is
not changed.

printf("%x\n",m );
return 0;
}
>
>
gives me :
$ gcc -Wall -g -o test test.c
test.c: In function `main':
test.c:4: warning: statement with no effect

Right.

$ ./test
fdff
>
>
whereas this works out properly :
>
#include<stdio. h>
int main(void){
unsigned int m=0xfdff;
printf("%x\n",~ m);

Here, you use the ~m expression as an argument to printf, so its value /is/
used, so it does have an effect on the program.

But m's value is unchanged. Add this line to your program:

printf("And the value of m is still %x\n", m);

If you want to change m's value, you will want to use some kind of
assignment statement.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999

http://www.cpax.org.uk

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