'Multiple conditions in ternary operators

First off, the question is "Write a Java program to find the smallest of three numbers using ternary operators."

Here's my code:

class questionNine
{
    public static void main(String args[])
    {
        int x = 1, y = 2, z = 3;
        int smallestNum;

        smallestNum = (x<y && x<z) ? x : (y<x && y<z) ? y : (z<y && z<x) ? z;
        System.out.println(smallestNum + " is the smallest of the three numbers.");
    }
}

I tried to use multiple conditions in the ternary operator but that doesn't work. I was absent a few days so I'm not really sure what to do and my teacher's phone is off. Any help?

javaoperator-keywordternary


Solution 1:[1]

Try

int min = x < y ? (x < z ? x : z) : (y < z ? y : z);

You can also remove the parenthesis:

int min = x < y ? x < z ? x : z : y < z ? y : z;

Solution 2:[2]

Since this is homework I'm not just going to give you the answer, but instead an algorithm so that you can work it out yourself.

First work out how to write min(x, y) using a single ternary operator.

Once you have that, change the following code for min(x, y, z) to use a ternary operator then substitute in the code for min(x, y) that you worked out from the previous step.

int min(x, y, z) {
    if (x <= y) {
        return min(x, z);
    } else {
        return min(y, z);
    }
}

Solution 3:[3]

You're testing for z, when you really don't need to. Your ternary operator must be of form cond ? ifTrue : ifFalse;

so if you have multiple conditions, you have this:

cond1? ifTrue1 : cond2? if True2 : ifFalse2;

If you understand this, don't look below. If you still need help, look below.

I also included a version that doesn't nest them that is clearer (assuming you don't need to have them nested. I sure would hope your assignment doesn't require you to nest them because that's pretty ugly!)

.

.

Here's what I come up with:

class QuestionNine
{
    public static void main(String args[])
    {
        smallest(1,2,3);
        smallest(4,3,2);
        smallest(1,1,1);
        smallest(5,4,5);
        smallest(0,0,1);
    }

    public static void smallest(int x, int y, int z) {
        // bugfix, thanks Mark! 
        //int smallestNum = (x<y && x<z) ? x : (y<x && y<z) ? y :  z;
        int smallestNum = (x<=y && x<=z) ? x : (y<=x && y<=z) ? y :  z;
        System.out.println(smallestNum + " is the smallest of the three numbers.");
    }

    public static void smallest2(int x, int y, int z) {
       int smallest = x < y ? x : y; // if they are equal, it doesn't matter which
       smallest = z < smallest ? z : smallest;
       System.out.println(smallest + " is the smallest of the three numbers.");
    }
}

Solution 4:[4]

public static int min(int x, int y, int z) {
    return x<y?Math.min(x, z):Math.min(y, z);           
}

public static void main(String[] args) {
    int smallestNum = min(10, 12, 15);
    System.out.println(smallestNum);
}

Solution 5:[5]

I know it's late. Just for information this also works:

int smallestNum = (x<y ? x : (x=y)) < z ? x : z

Solution 6:[6]

The last part: (z<y && z<x) ? z is missing a ':' :

(z<y && z<x) ? z : some_value;

Solution 7:[7]

My solution:

public static void main(String args[])
{
    int x = 1, y = 2, z = 3;
    int smallestNum = (x < y && x < z) ? x :
        (y < x && y < z) ? y :
        (z < y && z < x) ? z:y;
    System.out.println(smallestNum + " is the smallest of the three numbers.");
}

Solution 8:[8]

My contribution ... 

public static int getSmallestNumber( int x, int y, int z) {

     return x < y && x < z ? x : y < x && y < z ? y : z;
}

public static void main ( String ... args ) {

    System.out.println( getSmallestNumber( 123, 234, 345 ) );
}

Solution 9:[9]

public static void main(String args[]) {
    Scanner sc=new Scanner(System.in);
    System.out.println("enter the number and check heigest number of three number:-");
    int a=sc.nextInt();
    int b=sc.nextInt();
    int c=sc.nextInt();
    int max;
    max=(a>b&&a>c)?a:(b>a&&b>c)?b:(c>a&&c>b)?c:(a==b&&b==c)?a:0;
    
    System.out.println("the heigest number is:"+max);

Solution 10:[10]

This answer is coming in seven years late, so I'll just give you the code:

int smallestNumber = (x > y) ? 
            ((y > z) ? z : y) : 
            ((x > z) ? z : x);

The indentation should explain the code, which is simply a ternary that evaluates other ternaries on the initial condition x > y; /* the first ternary gets evaluated if that condition is true, else the second one gets evaluated. */

Solution 11:[11]

you missed the value after z var smallestNum = (x<y && x<z) ? x : (y<x && y<z) ? y : (z<y && z<x) ? z:0; Try this, it works

Solution 12:[12]

public class questionNine{
  public static void main(String args[]){
    int x = 1, y = 2, z = 3;
    int smallestNum =  (x<y && x<z) ? x : ((y < x && y<z) ? y : z);
    System.out.println(smallestNum);
  }
}

Solution 13:[13]

int min = (x<y)?((x<z)?x:z):((y<z)?y:z);

Solution 14:[14]

best way to do this is to make an example using if and else statement and then apply the ternary operators on it (q

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