help

[pdk123]

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

[Narue]

>can somebody tell me is there a problem in it ?
Assuming boolean is a typedef, there's nothing syntactically or semantically wrong with that function. Perhaps if you told us what it's supposed to do, we can tell you if it actually does it.

[pdk123]

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..

[Salem]

Your analysis is correct.

But whether there is anything 'wrong' depends on what it's supposed to do.
If for example is was a test to see if a number was >= 8, then it would fail miserably on negative numbers.

The name 'new' certainly doesn't help.

[pdk123]

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'

[Salem]

True.
So what's the indented function then?
If we don't know what it's supposed to do, how can we say whether it's right or wrong?

[pdk123]

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..

[stewie griffin]

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

[Colin Mac]

The only potential problem you can be sure about is that 'boolean' isn't a standard type.

[pdk123]

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'

[Salem]

So knowing what it does, you can write a spec which means there isn't a problem, or you can write a different spec where there is a problem.

I suppose pedantically, the parameter should be unsigned, if you're messing around with bits.

[ssharish2005]

Hmmmm that was good question, it need a 2 min think before we could answer that question. I never an interview question like that before. But that was good one.

I don't see any improvements when we could do to that code. Everything seem to be good. And I do think 7 needs to be there if you wanted check the last 3 bit of a to check if they are set.

And again it all depends upon the requirement!

ssharish

[Prabakar]

My guess:
1)The function checks whether the integer a is divisible by 8.
2)The use of new which is not an operator in c but in c++.

[pdk123]

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.

[pdk123]

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:

1 bool f( uint n )

{

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

}

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

2. uint f( uint n )

{

return –n & 7;

}

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

3. 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.
6. 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.

7. 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.

[Salem]

How about using code tags?

[pdk123]

how do i use it ?

[mitrmkar]

how do i use it ?

see here

[Salem]

It would certainly fail on -ve numbers which didn't use 2-s complement. Nothing in the C standard prevents other ways of storing integers .

[pdk123]

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