bitwise operators

[roman ziak]

hi,

I need to write a function that will check whether x is nonzero. Return 0
if x is = to 0 and 1 otherwise. The open operators I can use are ~ & ^ | +
<< >> and I am not allowed to use an loops.

My program so far looks like:

int isNonZero(int x) {


x = (1 << x) & x;
x = ~x;
x = x & 1;
return x;


}

I am not sure if standard defines the case when you are shifting left by
more than the bit size of integer.

But how about:

N-1
---
[ \ (x >> i) ] & 1
/
---
i=0

where N is the bit size of word, so for 4 bit machine the expression is:

( x + (x>>1) + (x>>2) + (x>>4) ) & 1

 

[roman ziak]

N-1

---
[ \ (x >> i) ] & 1
/
---
i=0

where N is the bit size of word, so for 4 bit machine the expression is:

( x + (x>>1) + (x>>2) + (x>>4) ) & 1

correction:

( x + (x>>1) + (x>>2) + (x>>3) ) & 1

 

[Chris Williams]

It must have a sign bit and at least 15 value bits, but it can be
one byte, or two, or three, or four, or five, or eight, or even
thirty-seven. I have worked on systems with 1-, 2-, and 4-byte
ints,

How does one get "a sign bit and at least 15 value bits" with 1-byte
ints? Or is this a platform with 16-bit bytes?

and I know of at least one system with 8-bit bytes.

8-bit bytes? x86?

But back to the drawing board I guess for my bit-count determiner =)

-Chris

 

[roman ziak]

N-1
---
[ \ (x >> i) ] & 1
/
---
i=0

where N is the bit size of word, so for 4 bit machine the expression is:

( x + (x>>1) + (x>>2) + (x>>4) ) & 1

correction:

( x + (x>>1) + (x>>2) + (x>>3) ) & 1

sorry, one more correction:

( x | (x>>1) | (x>>2) | (x>>3) ) & 1

this should be the last

 

[infobahn]

How does one get "a sign bit and at least 15 value bits" with 1-byte
ints? Or is this a platform with 16-bit bytes?

32, in my case.

8-bit bytes? x86?

Bigtime typo. I meant 8-byte ints.

 

[Randy Howard]

This has it. That's the first thing I thought of. Any excuse to stack up
operators.

If you don't like that then, try return !!!!x; instead


It's best not to use tricks though, readability is paramount.

also, this will suffice, return x ? 1 : 0 ;

Why are doing obvious homework assignments, no matter how trivial?

 

[infobahn]

Why are doing obvious homework assignments, no matter how trivial?

I don't think we did, unless you count "this homework assignment
is impossible" as "doing" the assignment.

 

[Mark F. Haigh]

hi,

I need to write a function that will check whether x is nonzero. Return 0
if x is = to 0 and 1 otherwise. The open operators I can use are ~ & ^ | +
<< >> and I am not allowed to use an loops.

<snip>


#include <stdio.h>
#include <stdint.h>

int func(uint32_t x)
{
return ~((x - 1) & ~x & 0x80000000u) >> 31;
}


I think this should work. Needs C99 for the exact width unsigned
integer type.


Mark F. Haigh
(e-mail address removed)

 

[infobahn]

<snip>

#include <stdio.h>
#include <stdint.h>

int func(uint32_t x)
{
return ~((x - 1) & ~x & 0x80000000u) >> 31;
}

I think this should work. Needs C99 for the exact width unsigned
integer type.

Even if we take C99 for granted (and that's a big if), what if
UINT_MAX < 0x80000000u ?

 

[Michael Mair]

C99 does not guarantee the exact width integer types.

Even if we take C99 for granted (and that's a big if), what if
UINT_MAX < 0x80000000u ?

You know it, I know it, so let's just sing it together:
0x80000000ul


Cheers
Michael

 

[Mark F. Haigh]

Even if we take C99 for granted (and that's a big if), what if
UINT_MAX < 0x80000000u ?

Huh? With C99, if 0x80000000u is not representable as an unsigned int,
it becomes a unsigned long.

But it's not really production code we're talking about here, is it?


Mark F. Haigh
(e-mail address removed)

 

[infobahn]

C99 does not guarantee the exact width integer types.

And in any case, the OP specified int, not uint32_t