Display '>', '<', or '=' in between numbers

[crumby1]

Given five integers as input, display each in the order entered by the user with the character '>', '<', or '=' in between each pair of numbers that accurately represents their relationship.

How do i get the greater than, less than, equal to signs to switch? I am not allowed to use logical operators, relational operators, or selection constructs.

Here is what the program should look like:

Example Execution #1:
Enter the five integer values: 4 8 9 0 3
4 < 8 < 9 > 0 < 3

Example Execution #2:
Enter the five integer values: 2 2 3 3 2
2 = 2 < 3 = 3 > 2


Thanks in advance!

[JLBorges]

#include <iostream>

int main()
{
    const int n = 5 ;
    int numbers[n] ;

    // read in n numbers
    for( int i = 0 ; i < n ; ++i ) std::cout << "? " && std::cin >> numbers[i] ;

    std::cout << '\n' << numbers[0] ; // print the first number

    // create an array of symbols: symbols[0] == '>' symbols[1] == '=' symbols[2] == '<'
    const char symbol[] = { '>', '=', '<' };

    // for the next n-1 numbers
    for( int i = 1 ; i < n ; ++i )
    {
        const int previous_number = numbers[i-1] ;
        const int this_number = numbers[i] ;

        // not allowed to use logical operators, relational operators, or selection constructs.
        // compute the difference between this number and the previous number
        int diff = this_number - previous_number ;
        // diff is negative if previous_number > this_number,
        //         zero if previous_number == this_number,
        //         positive if previous_number > this_number

        // if diff is zero, leave it unchanged
        // if diff is non-zero, make it an odd number
        diff += bool(diff) + (diff%2) ;
        // note: bool(diff) + (diff%2) == 0+0 == 0 if diff is zero
        // note: bool(diff) + (diff%2) == 1-1 == 0 if diff is an odd negative number
        // note: bool(diff) + (diff%2) == 1+1 == 2 if diff is an odd positive number
        // note: bool(diff) + (diff%2) == 1-0 == 1 if diff is an even negative number
        // note: bool(diff) + (diff%2) == 1+0 == 1 if diff is an even positive number

        const int index = (diff%2) + 1 ; // index == zero if diff is negative,
                                         // index == 1 if diff is zero,
                                         // index == 2 if diff is positive

        // print the appropriate symbol followed by this number
        std::cout << ' ' << symbol[index] << ' ' << this_number ;
    }

    std::cout << '\n' ; // finally print a newline
}

http://coliru.stacked-crooked.com/a/807c791917d36b0b

[dhayden]

If you had int sgn(int x) that returned -1, 0, or 1 depending on whether x was negative, zero, or positive, then JLBorges's code could be simplified:

const char symbol[] = { '<', '=', '>' };
...
std::cout << ' ' << symbol[1+sgn(diff)] << ' ' << this_number ;

Poking around I found this nice sequence for generating sgn: (x > 0) - (x < 0):
http://stackoverflow.com/questions/1903954/is-there-a-standard-sign-function-signum-sgn-in-c-c

This can be done in just a few instructions on most CPUs because (x<0) can be computed by shifting the sign bit down to bit 0. To compute x>0 you just do (-x) < 0.