• C

Detect set bits

I need to determine if bits 0, 1, 10 and 11 are set
in unsigned short that is passed to me.

I wrote the code, but there's no way for me to test!
Need to make sure I'm getting it right.
Value passed to me will something like 3515, or so I was told.

Plz help, urgent.
jd9288Asked:
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Kent OlsenConnect With a Mentor Data Warehouse Architect / DBACommented:
You can test this easy enough.

#define BITMASK ((1 << 0) | (1 << 1) | (1 << 10) | (1 << 11))

unsigned int TestBit (unsigned integer Value)
{
  unsigned int Result;

  Result = Value & BITMASK;
  printf ("Test Value = %8.8x, Mask=%8.8x, Result=%8.8x\n", Value, BITMASK, Result);
}

If the displayed value for Result is the same as for BITMASK, then all bits in the bitmask were set in the original word.


Kent
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jd9288Author Commented:
BTW, the code I have is this:

ReadBit(int dat, int bitno)
{

     int dattest;

     //
     // shift desired bit into rightmost position
     //
     dattest = (dat >> bitno);

     return (dattest & 1);

}
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joghurtConnect With a Mentor Commented:
unsigned short value = ...;

if ((value & (1 << 0)) != 0) {
  // bit 0 is set
}
if ((value & (1 << 1)) != 0) {
  // bit 1 is set
}
etc.

For bits 0, 1, 10 AND 11 (at the same time) you'd try (1 << 0) | (1 << 1) | (1 << 10) | (1 << 11).
You may be surprised but the above expression gives me 3075 (or 0x0C03) instead of 3515.
So for checking what you want you should write

if ((value & 0x0C03) == 0x0C03) {
// All the required bits are set.
}

OR

if ((value & 0x0C03) != 0) {
// At least one of the required bits are set.
}
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Kent OlsenData Warehouse Architect / DBACommented:
Hi jd9288,

#define BITMASK ((1 << 0) | (1 << 1) | (1 << 10) | 1 << 11))


if (Value == BITMASK)  /*  All bits in bitmask are set, no others are set  */

if (Value & BITMASK)   /*  All bits in bitmask are set, others may be set  */

if (Value & BITMASK)   /*  Any bit in bitmask is set  */



Good Luck,
Kent
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Kent OlsenData Warehouse Architect / DBACommented:
Whoops...

#define BITMASK ((1 << 0) | (1 << 1) | (1 << 10) | (1 << 11))


:-)

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Kent OlsenData Warehouse Architect / DBACommented:
Man, what a day...

#define BITMASK ((1 << 0) | (1 << 1) | (1 << 10) | (1 << 11))


if (Value == BITMASK)  /*  All bits in bitmask are set, no others are set  */

if ((Value & BITMASK) == BITMASK)  /*  All bits in bitmask are set, others may be set  */

if (Value & BITMASK)   /*  Any bit in bitmask is set  */


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bkfirebirdConnect With a Mentor Commented:
>> I need to determine if bits 0, 1, 10 and 11 are set
does this mean you want  to check whether the four least significant bits of the number are set?
your ReadBit function looks fine ... you can check if a particular bit is set by calling .....

if (ReadBit(num, 1)) // bit is set
// do something

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jd9288Author Commented:
Thank you all.
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