K&R Exercise 3-4: Negative Numbers Represented In Binary

Question

I'm having a hard time understanding this exercise:

In a two's complement number representation, our version of itoa does not handle the largest negative number, that is, the value of n equal to -(2^(wordsize-1)). Explain why not. Modify it to print that value correctly, regardless of the machine on which it runs.

Here is what the itoa originally looks like:

void reverse(char s[], int n)
{
  int toSwap;
  int end = n-1;
  int begin = 0;

  while(begin <= end) // Swap the array in place starting from both ends.
    {
      toSwap = s[begin];
      s[begin] = s[end];
      s[end] = toSwap;

      --end;
      ++begin;
    }
}

// Converts an integer to a character string.
void itoa(int n, char s[])
{
  int i, sign;
  if ((sign = n) < 0)
    n = -n;
  i = 0;
  do
    {
      s[i++] = n % 10 + '0';
    } while ((n /= 10) > 0);

  if (sign < 0)
    s[i++] = '-';
  s[i] = '\0';
  reverse(s, i);
}

I found this answer, but I don't understand the explanation: http://www.stevenscs.com/programs/KR/$progs/KR-EX3-04.html

Because the absolute value of the largest negative number a word can hold is greater than that of the largest positive number, the statement early in iota that sets positive a negative number corrupts its value.

Are they saying that negative numbers contain more bits because of the sign than a positive number which has no sign? Why would multiplying by -1 affect how the large negative number is stored?

Why it takes more bits to store a signed number? Because of the bit sign.... — Ryan B., Aug 31 '16 at 20:44
@RyanB because the largest negative number is 1 followed by all 0 (-(2^(wordsize-1))), while the largest positive number is 0 followed by all 1 (2^(wordsize-1)-1). — Davide Visentin, Aug 31 '16 at 20:47

score 1 · Accepted Answer · answered Aug 31 '16 at 20:59

In two's complement representation, the range of values you can represent is -2^n-1 to 2^n-1-1. Thus, with 8 bits, you can represent values in the range -128 to 127. That's what's meant by the phrase, "the largest negative number a word can hold is greater than that of the largest positive number."

Illustrating with just 3 bits to make it clearer:

Value    Bits
-----    ----
    0    000
    1    001
    2    010
    3    011
   -4    100
   -3    101
   -2    110
   -1    111

With 3 bits, there's no way we can represent a positive 4 in two's complement, so n = -n; won't give us the result we expect¹. That's why the original atoi implementation above can't deal with INT_MIN.

^{Behavior on signed integer overflow is undefined, meaning that there's no fixed result.}

score 0 · Answer 2 · answered Aug 31 '16 at 20:45

0

The problem is that, if n is the largest negative number, when you do n=-n you obtain 0, bacause you cannot represent a positive number that big.

A solution can be to hold the positive number in a long integer.

answered Aug 31 '16 at 20:45

Davide Visentin

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K&R Exercise 3-4: Negative Numbers Represented In Binary

2 Answers2