检查输入是否为最大的2的补码整数

Question

/*
 * isTmax - returns 1 if x is the maximum, two's complement number,
 *     and 0 otherwise 
 *   Legal ops: ! ~ & ^ | +
 *   Max ops: 10
 *   Rating: 1
 */
int isTmax (int x) 
{
    int t = x + 1;
    return !(t + t) ^ !t;
}  

If x is the largest 2's complement number then 2*t overflows to 0 but its also 0 if x is -1 so we xor it with !t which evaluates to 1. So we must get 1 but somehow the output is 0.

Answer 1

This code is fundamentally wrong. Signed overflow is undefined behavior and not guaranteed to give any form of deterministic result.

To check if a number is the "largest possible", you would rather compare it with INT_MAX from limits.h. Or if you will, compare it with (int) ((1u<<31)-1) .

Answer 2

int isTmax (int x) {
    return !(~(x + x + 1) | !(~x));
}

This works as (x+x+1) becomes -1 if x is INT_MAX and all that's left is to check if x is not -1.

Answer 3

If constants are allowed, you can do

int isTmax (int x) 
{
   return !(((unsigned) x) ^ 0x7fffffff);
}

Otherwise, you can do

int isTmax (int x) 
{
unsigned nx = ~(unsigned) x;
unsigned r = !(nx + nx) ^ !nx;
}