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main()
{
int i;
clrscr();
printf("%d", &i)+1;
scanf("%d", i)-1;
}

whats the output n why???
Note that there is & which means it probably prints the address of
i...n my guess is that + 1 has no effect on the output...
n regarding scanf - there is no & symbol...so still it will accept a
value but not store it in i n my guess is that -1 has no effect on the
code....
any comments....

wrote:
> main()

int main (void)

> {
> int i;
> clrscr();

Not a standard function.

> printf("%d", &i)+1;

printf is expecting an int, you are passing a pointer to int, this is
undefined behavior. Try printf("%d\n", i);
You need to #include <stdio.h> before calling printf.
You need to initialize i before using it's value, what you are doing is
undefined behavior.

> scanf("%d", i)-1;

scanf is expecting a pointer to int, you are passing an int. Try
scanf("%d", &i);
You need to #include <stdio.h> before calling scanf.

> }
>
> whats the output n why???

You have several examples of undefined behavior in your program, this
means anything goes, you shouldn't expect anything in particular.

> Note that there is & which means it probably prints the address of
> i...n my guess is that + 1 has no effect on the output...

If you want to print the address, cast the address of i to a pointer to
void and use the %p conversion specifier:

printf("%p\n", (void *)&i);

printf returns an integer value. In your statement 1 is added to the
this value and the result is discarded. The compiler could legally
just remove the +1 as an optimization since your program wouldn't know
the difference, it doesn't use the result in any way.

Robert Gamble