Smallest of 3 numbers without using Comparator operator.

[Nethra Nayak]

How to Write a C program to find the smallest of three integers, without using any of the comparision operators.?

[JosAH]

How to Write a C program to find the smallest of three integers, without using any of the comparision operators.?

Why do you want to do that? Is it a riddle from your textbook? If so, this is not
a do-my-homework service; first give it a try yourself.

kind regards,

Jos (moderator)

hint: check whether or not the sign bit of (a-b) is set.

[whodgson]

say 3 nums are n1,n2 & n3
say n1=3,n2=15 & n3=9
ni*n2==45...... ..........(1)
n1*n3==27...... .........(2)
n2*n3==135..... ........(3)
the smallest num is common to expression 1 and 2 which is n1 or 3
how rediculous!

[Ganon11]

Yes, but to determine which result was smallest, you'd have to use...a comparison operator.

[JosAH]

As I wrote before: use the sign bit of an expression. Something like this:

#define ALL ((unsigned int)~0)
#define SGN (ALL^(ALL>>1))
#define LESS(X,Y) (((X)-(Y))&SGN)

LESS(x, y) will be true (not 0) when x < y. This trickery-dickery only works for
ints.

kind regards,

Jos

[whodgson]

Gannon11 says:
Yes, but to determine which result was smallest, you'd have to use...a comparison operator.

Why couldn't the 3 products be printed out as shown and then
cout<<"Which int: ";
cin>>numsmall;
user chooses n1 or n2 or n3 on the basis of which n is common to two smallest products
cout<<"The smallest int is: "<<numsmall ;
QED but agreed <<<<<< than elegant

[Alan Lopez]

Let your three integers be:

int num1, num2, num3;

//starts computing smallest number

if ( num1 < num2 ) {
if ( num1 < num3 ) {
printf( "Smallest if is %d\n", num1 );
}
}
if ( num2 < num1 ) {
if ( num2 < num3 ) {
printf( "Smallest if is %d\n", num2 );
}
}
if ( num3 < num1 ) {
if ( num3 < num2 ) {
printf( "Smallest if is %d\n", num3 );
}
}