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1. Are bit fields portable?

Bit fields are not portable. Because bit fields cannot span machine words, and because the number of bits
in a machine word is different on different machines, a particular program using bit fields might not even
compile on a particular machine.

Assuming that your program does compile, the order in which bits are assigned to bit fields is not defined.
Therefore, different compilers, or even different versions of the same compiler, could produce code that
would not work properly on data generated by compiled older code. Stay away from using bit fields, except
in cases in which the machine can directly address bits in memory and the compiler can generate code to take
advantage of it and the increase in speed to be gained would be essential to the operation of the program.

2. How are 16- and 32-bit numbers stored?

A 16-bit number takes two bytes of storage, a most significant byte and a least significant byte. If you write the 16-bit number on paper, you would start with the most significant byte and end with the least significant byte. There is no convention for which order to store
them in memory, however.

Let's call the most significant byte M and the least significant byte L. There are two possible ways to store
these bytes in memory. You could store M first, followed by L, or L first, followed by M. Storing byte M first
in memory is called "forward" or "big-endian" byte ordering. The term big endian comes from the fact that
the "big end" of the number comes first, and it is also a reference to the book Gulliver's Travels, in which the
term refers to people who eat their boiled eggs with the big end on top.

Storing byte L first is called "reverse" or "little-endian" byte ordering. Most machines store data in a big-
endian format. Intel CPUs store data in a little-endian format, however, which can be confusing when
someone is trying to connect an Intel microprocessor-based machine to anything else.

A 32-bit number takes four bytes of storage. Let's call them Mm, Ml, Lm, and  Ll in decreasing order of
significance. There are 4! (4 factorial, or 24) different ways in which these bytes can be ordered. Over the
years, computer designers have used just about all 24 ways. The most popular two ways in use today, however,
are (Mm, Ml, Lm, Ll), which is big-endian, and (Ll, Lm, Ml, Mm), which is little-endian. As with 16-bit numbers,
most machines store 32-bit numbers in a big-endian format, but Intel machines store 32-bit numbers in a
little-endian format.

3. Is it better to bitshift a value than to multiply by 2?

Any decent optimizing compiler will generate the same code no matter which way you write it. Use whichever
form is more readable in the context in which it appears. The following program's assembler code can be
viewed with a tool such as CODEVIEW on DOS/Windows or the disassembler (usually called "dis") on
UNIX machines:

Example: Multiplying by 2 and shifting left by 1 are often the same.

void main()
{
  unsigned int test_nbr = 300;
  test_nbr *= 2;
  test_nbr = 300;
  test_nbr <<= 1;
}

4. What is meant by "bit masking"?

Bit masking means selecting only certain bits from byte(s) that might have many bits set. To examine some
bits of a byte, the byte is bitwise "ANDed" with a mask that is a number consisting of only those bits of interest.
For instance, to look at the one's digit (rightmost digit) of the variable flags, you bitwise AND it with a mask
of one (the bitwise AND operator in C is &):

flags & 1;

To set the bits of interest, the number is bitwise "ORed" with the bit mask (the bitwise OR operator in C is
|). For instance, you could set the one's digit of flags like so:

flags = flags | 1;

Or, equivalently, you could set it like this:

flags |= 1;

To clear the bits of interest, the number is bitwise ANDed with the one's complement of the bit mask. The
"one's complement" of a number is the number with all its one bits changed to zeros and all its zero bits
changed to ones. The one's complement operator in C is ~. For instance, you could clear the one's digit of
flags like so:

flags = flags & ~1;

Or, equivalently, you could clear it like this:

flags &= ~1;

Sometimes it is easier to use macros to manipulate flag values.

Example Program : Macros that make manipulating flags easier.

/* Bit Masking */
/* Bit masking can be used to switch a character
   between lowercase and uppercase */
#define BIT_POS(N)            ( 1U << (N) )
#define SET_FLAG(N, F)        ( (N) |= (F) )
#define CLR_FLAG(N, F)        ( (N) &= -(F) )
#define TST_FLAG(N, F)        ( (N) & (F) )
#define BIT_RANGE(N, M)       ( BIT_POS((M)+1 - (N))-1 << (N) )
#define BIT_SHIFTL(B, N)      ( (unsigned)(B) << (N) )
#define BIT_SHIFTR(B, N)      ( (unsigned)(B) >> (N) )
#define SET_MFLAG(N, F, V)    ( CLR_FLAG(N, F), SET_FLAG(N, V) )
#define CLR_MFLAG(N, F)       ( (N) &= ~(F) )
#define GET_MFLAG(N, F)       ( (N) & (F) )
#include <stdio.h>
void main()
{
  unsigned char ascii_char = 'A';        /*  char = 8 bits only */
  int test_nbr = 10;
  printf("Starting character = %c\n", ascii_char);
  /*  The 5th bit position determines if the character is
      uppercase or lowercase.
      5th bit = 0  - Uppercase
      5th bit = 1  - Lowercase      */
  printf("\nTurn 5th bit on = %c\n", SET_FLAG(ascii_char, BIT_POS(5)) );
  printf("Turn 5th bit off = %c\n\n", CLR_FLAG(ascii_char, BIT_POS(5)) );
  printf("Look at shifting bits\n");
  printf("=====================\n");
  printf("Current value = %d\n", test_nbr);
  printf("Shifting one position left = %d\n",
         test_nbr = BIT_SHIFTL(test_nbr, 1) );
  printf("Shifting two positions right = %d\n",
         BIT_SHIFTR(test_nbr, 2) );
}

BIT_POS(N) takes an integer N and returns a bit mask corresponding to that single bit position (BIT_POS(0)
returns a bit mask for the one's digit, BIT_POS(1) returns a bit mask for the two's digit, and so on). So instead
of writing

#define A_FLAG  4096

#define B_FLAG  8192

you can write

#define A_FLAG  BIT_POS(12)

#define B_FLAG  BIT_POS(13)

which is less prone to errors.

The  SET_FLAG(N, F) macro sets the bit at position F of variable N. Its opposite is  CLR_FLAG(N, F), which clears
the bit at position F of variable N. Finally, TST_FLAG(N, F) can be used to test the value of the bit at position
F of variable N, as in

if (TST_FLAG(flags, A_FLAG))
        /* do something */;

The macro  BIT_RANGE(N, M) produces a bit mask corresponding to bit positions N  through M, inclusive. With
this macro, instead of writing

#define FIRST_OCTAL_DIGIT  7       /* 111 */

#define SECOND_OCTAL_DIGIT 56      /* 111000 */

you can write

#define FIRST_OCTAL_DIGIT  BIT_RANGE(0, 2)    /* 111 */

#define SECOND_OCTAL_DIGIT BIT_RANGE(3, 5)    /* 111000 */

which more clearly indicates which bits are meant.

The macro BIT_SHIFT(B, N) can be used to shift value B into the proper bit range (starting with bit N). For
instance, if you had a flag called C that could take on one of five possible colors, the colors might be defined
like this:

#define C_FLAG          BIT_RANGE(8, 10)      /* 11100000000 */
/* here are all the values the C flag can take on */
#define C_BLACK         BIT_SHIFTL(0, 8)       /* 00000000000 */
#define C_RED           BIT_SHIFTL(1, 8)       /* 00100000000 */
#define C_GREEN         BIT_SHIFTL(2, 8)       /* 01000000000 */
#define C_BLUE          BIT_SHIFTL(3, 8)       /* 01100000000 */
#define C_WHITE         BIT_SHIFTL(4, 8)       /* 10000000000 */
#define C_ZERO          C_BLACK
#define C_LARGEST       C_WHITE
/* A truly paranoid programmer might do this */
#if C_LARGEST > C_FLAG
        Cause an error message. The flag C_FLAG is not
        big enough to hold all its possible values.
#endif /* C_LARGEST > C_FLAG */

The macro SET_MFLAG(N, F, V) sets flag F in variable N to the value V. The macro CLR_MFLAG(N, F) is identical
to CLR_FLAG(N, F), except the name is changed so that all the operations on multibit flags have a similar
naming convention. The macro GET_MFLAG(N, F) gets the value of flag F in variable N, so it can be tested,
as in

if (GET_MFLAG(flags, C_FLAG) == C_BLUE)
        /* do something */;

5. What is meant by high-order and low-order bytes?

We generally write numbers from left to right, with the most significant digit first. To understand what is
meant by the "significance" of a digit, think of how much happier you would be if the first digit of your
paycheck was increased by one compared to the last digit being increased by one.

The bits in a byte of computer memory can be considered digits of a number written in base 2. That means
the least significant bit represents one, the next bit represents 2'1, or 2, the next bit represents 2'2'1, or
4, and so on. If you consider two bytes of memory as representing a single 16-bit number, one byte will hold
the least significant 8 bits, and the other will hold the most significant 8 bits. Figure shows the bits
arranged into two bytes. The byte holding the least significant 8 bits is called the least significant byte, or
low-order byte. The byte containing the most significant 8 bits is the most significant byte, or high-
order byte.

6. What is the most efficient way to store flag values?

A flag is a value used to make a decision between two or more options in the execution of a program. For
instance, the /w flag on the MS-DOS dir command causes the command to display filenames in several
columns across the screen instead of displaying them one per line. In which a flag is used to indicate which of two possible types is held in a union.
Because a flag has a small number of values (often only two), it is tempting to save memory space by not
storing each flag in its own int or char.

Efficiency in this case is a tradeoff between size and speed. The most memory-space efficient way to store
a flag value is as single bits or groups of bits just large enough to hold all the possible values. This is because
most computers cannot address individual bits in memory, so the bit or bits of interest must be extracted from
the bytes that contain it.

The most time-efficient way to store flag values is to keep each in its own integer variable. Unfortunately,
this method can waste up to 31 bits of a 32-bit variable, which can lead to very inefficient use of memory.
If there are only a few flags, it doesn't matter how they are stored. If there are many flags, it might be
advantageous to store them packed in an array of characters or integers. They must then be extracted by a
process called bit masking, in which unwanted bits are removed from the ones of interest.

Sometimes it is possible to combine a flag with another value to save space. It might be possible to use high-
order bits of integers that have values smaller than what an integer can hold. Another possibility is that some
data is always a multiple of 2 or 4, so the low-order bits can be used to store a flag.