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int v; // we want to find the absolute value of v
unsigned int r; // the result goes here
int const mask = v >> sizeof(int) * CHAR_BIT - 1;

r = (v + mask) ^ mask;

Reference: graphics.stanford.edu/~seander/bithacks.html#IntegerAbs

You can also watch this video to understand this solution
youtube.com/watch?v=zJGts7hOX9s

use 2's complement conversion(complemeent and add one), if no is in 2's complement, utilize the first bit of number and
the xor operation(with 00000...00 or 111111..111) which either complements or keep the number same

Using the property of multiplication: multiplying either two negative or either two positive numbers the result is always positive.

int x=-10;

abs= sqrt((float)(x*x));

the best one line solution i can think of:
//a be the no. whose abs has to befind...

cout<<(a^(-(0>a)))+(0>a);
//tested worked well for signed integers...:)

while (v < 0) {
   return -v;
}
return v;

What about try-catch?

public int abs(int n)
{
	try{
		int a = int.MaxValue + n;
		return -n;	//if it gets here, n is negative
	}
	catch{
		return n;	//if gets here, n is positive
	}
}

#include <stdio.h>
#include <stdlib.h>

int
main(void)
{
    int array[] = {70, 30};
    printf("largest: %d\n",array[!(array[0] > array[1])]);
    return 0;
}

basic assumption is incorrect here.
abs(-v) = v and abs(v) = v
you can refer any maths book for that.

abs(x) : if (x <0) then return 2's complement of x else return x


result = ( (x<0) & (~x+1)) | ( (x>0) & x )

means if x < 0 then find 2's complement of negative number and return
if x is positive return x anyways

x=(x>0)*x+(x<0)*-x;

x=(x>0)*x+(x<0)*-x;

n & (n – 1) == 0 //include 0
n & !(n & (n – 1))

int myabs(int a){
    int mask = a >> (sizeof(int)*8-1);
    a = (a+mask)^mask;
    return a;
}

(1&&(a&(1<<31)))*(-a) + (1&&((-a)&(1<<31)))*(a);