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cbooleanunary-function

question about C pointer on boolean value,

int bar(int *arr, size_t n)
{
    int sum = 0, i;
    for (i = n; i > 0; i--)
    {
        sum += !arr[i - 1];
    }
    return ~sum + 1;
}

I have come across this code but don't quite understand sum += !arr[i - 1];: what is the effect of !(NOT) applied to the pointer of an array? Also, what is the effect of ~ before sum?

Solution 
  • sum += !arr[i - 1];

    The ! is the Logical Negation operator. It is not applied on a pointer as you mention, but on the value arr[i-1]. If arr[i-1] ==0 the result is 1 otherwise the result is 0.

    ~sum + 1;

    ~ is the Bitwise NOT operator It will invert all the binary bits of sum. It is sometimes also called the ones complement

    The result of ~sum +1 is the same as taking the two's complement of sum, which is equal to the negative of sum. If sum is 5 it will return -5

    Some more explanation on the Logical Operators

    When the Logical operators (Logical AND, OR, NOT) are applied on a variable, it only checks the logical state of the variable. i.e. this is whether this is 0 or non 0 Non zero can take any value e.g. 5, 10, -5 etc.

    So, if you apply !0 you get the value of 1. For any other value e.g. !5 the answer is 0.

    From C99 6.5.3.5

    The result of the logical negation operator ! is 0 if the value of its operand compares unequal to 0, 1 if the value of its operand compares equal to 0. The result has type int. The expression !E is equivalent to (0==E)