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Comparing two integers without any comparison

Asked 2009-01-24T22:37:57.550
12

Is it possible to find the greatest of two integers without any comparison? I found some solutions: 

if(!(a/b)) // if a is less than b then division result will be zero.
{
    cout << " b is greater than a";
}
else if (!(a-b)) // we know a is greater than or equal to b now.  check whether they are equal.
{
    cout << "a and b are equal";
}
else
    cout << "a is greater than b";

But if(c) or if(!c) is a comparison to zero. In addition it doesn't work for negative numbers. In fact I need a solution that avoids any if statement. Instead I should use switch statements and arithmetic operators. ThanX.

c++ if-statement comparison
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2 Answers

1

As a pointless exercise, here's a way of implementing a cond function - to serve the purpose of if, supposing it (and switch, and ?:) had somehow disappeared from the language, and you're using C++0x.

void cond(bool expr, std::function<void ()> ifTrue, std::function<void ()> ifFalse)
{
    std::function<void ()> choices[2] = { ifTrue, ifFalse };
    choices[expr == false]();
}

e.g.

cond(x > y,
    /*then*/ [] { std::cout << "x is greater than y"; },
    /*else*/ [] { std::cout << "x is not greater than y"; });

Like I say, pointless.

answered 2009-01-25T00:43:41.570
-2
void greater(int a, int b) {
    int c = a - b;
    switch(c) {
        case 0:
            cout << "a and b are equal" << endl;
            break;
        default:
            int d = c & (1<<31);
            switch(d) {
                case 0:
                    cout << "a is bigger than b" << endl;
                    break;
                default:
                    cout << "a is less than b" << endl;
            }
    }
}
answered 2009-01-24T23:07:07.147

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