Posts by pdk123

yes, i.e TRUE, for other representations of -ve numbers it will fail.

how do i use it ?

now i got some part of this answer:
i have couple of other questions in the same form.

which i gave some answers, May be u can correct me , if i was wrong:

bool f( uint n )

{

  return (n & (n-1)) == 0;

}

My answer was : if n is powers of 2 then return 1 else '0'

uint f( uint n )

{

  return –n & 7;

}

My answer: last 3 (LSB) bits of the 2's complement of the input number .

int f( int n, int l, int r )

{

  return (n << l) >> r;

}

my Answer : n is bit shifted left by l bits and result is again right shifted by n bits.

void fn(long* p1, long* p2)

{

      register int x = *p1;

      register int y = *p2;

      x ^= y;

      y ^= x;

      x ^= y;

      *p1 = x;

      *p2 = y;

}

my answer : swaps the two inputs x and y.

void send(int count, short *to, short *from)

{

    /* count > 0 assumed */

    register n = (count + 7) / 8; 

    switch (count % 8)

    {

    case 0:        do {  *to = *from++;

    case 7:              *to = *from++;

    case 6:              *to = *from++;

    case 5:              *to = *from++;

    case 4:              *to = *from++;

    case 3:              *to = *from++;

    case 2:              *to = *from++;

    case 1:              *to = *from++;

                   } while (--n > 0);

    }

}

My answer : this expression copies the counth element of 'from' array to the first element of the 'to' array.

yes, this looks like the correct answer.

I was thinking that for -ve numbers this may fail. But this even checks for -ve numbers stored in 2's complenent form correctly i guess.

Then i donot see any potential problem in the function. except that function returns 1 for '0' also and all numbers (-ve and +ve ) divisible by 8.

This function resave number, if the number is 0 its return true else its return false.
The Boolean " a & 7 ;" says if a != 0 and 7 != 0 then 1 else 0
Has you see the 7 is not needed

--------------------------------

when the number is lets say '8'. it still returns '1' .. i.e basically numbers for which lsb 3 bits are not set. it will return '1'

even i was not sure. because in the test, i was supposed to tell what the function is intended to do and any potential problems !!!.
i could not exactly figure out this .. i.e why any pointers to this will be helpful.

as pointed out, i was also thinking whether it will cause any problems for -ve numbers .. but without any potential soln..

thankyou for the reply.
but i am not sure whether this logic will work ( i.e to check whether the number is >=8) .. as lets say for eg +9 , the program still returns '0'

it was asked for me in the test. where they asked me
1.what the code does and
2. the potential problem with that.

i answered, If any of the last 3 lsb bits of the number is set to '1' return '0'. Else return 1. and i could not see any potential problem in that..

following is a simple embedded c piece of code, can somebody tell me is there a problem in it ?

boolean new(int a)
{
      return !(a & 7);
}

i am new to embedded programming. It will be great help, any pointers to this ...

thanks
pdk