boy boy - 1 year ago 116
C Question

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

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;


// Converts an integer to a character string.
void itoa(int n, char s[])
int i, sign;
if ((sign = n) < 0)
n = -n;
i = 0;
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:$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?

Answer Source

In two's complement representation, the range of values you can represent is -2n-1 to 2n-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.

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

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