问题
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?
回答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;
回答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);
}
}
回答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.");
}
}
回答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);
}
回答5:
I know it's late. Just for information this also works:
int smallestNum = (x<y ? x : (x=y)) < z ? x : z
回答6:
The last part: (z<y && z<x) ? z
is missing a ':' :
(z<y && z<x) ? z : some_value;
回答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.");
}
回答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 ) );
}
回答9:
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. */
回答10:
int min = (x<y)?((x<z)?x:z):((y<z)?y:z);
回答11:
best way to do this is to make an example using if and else statement and then apply the ternary operators on it (q
来源:https://stackoverflow.com/questions/4986654/multiple-conditions-in-ternary-operators