How do I restrict the range permitted for the int argument of a method

Question

Say I have a method:

public void method(int i) {
    this.value =+ i;
}

How can I restrict i to a value greater than x? I understand this can be done outside the method but I would prefer to condition it in the constructor. Would it be something like:

public void method(int i; i > x){...};

Sorry for not testing this myself, I came up with it as I wrote my query.

As far as i know you cannot do in it the method definition but you can do it in the body. — StackFlowed, Nov 25 '14 at 18:52
You can check before the call, check inside the call, use some form of annotation-based pre-/post-condition checking, delegate to the underlying method using a validation/assertion method, etc. — Dave Newton, Nov 25 '14 at 18:54
Personally, rather than just throwing something up on SO, I'd become familiar with the language syntax. There are many documents that cover what is and isn't valid Java. — Dave Newton, Nov 25 '14 at 18:57
See also http://stackoverflow.com/questions/6816298/is-there-an-beautiful-way-to-assert-pre-conditions-in-java-methods — Alexis C., Nov 25 '14 at 19:00
Are you trying to use the compound assignment "+="? Because if you do, you have it backwards in your code. — Kirill Rakhman, Nov 29 '14 at 14:32

score 9 · Accepted Answer · answered Nov 25 '14 at 18:52

There is no syntax in Java that can restrict the range of a parameter. If it takes an int, then all possible values of int are legal to pass.

You can however test the value with code in the method and throw an IllegalArgumentException if i is not greater than x.

JB Nizet · Answer 2 · 2014-11-25T18:59:44.167

7

You document the method, and throw an exception if the caller doesn't respect the contract.

/**
 * Does something cool.
 * @param i a value that must be > x
 * @throws IllegalArgumentException if i <= x
 */
public void method(int i) {
    if (i <= x) {
        throw new IllegalArgumentException("i must be > x");
    }
    this.value += i;
}

edited Nov 25 '14 at 18:59

answered Nov 25 '14 at 18:54

JB Nizet

657,433
87
1,179
1,226

I would do it throw new IllegalArgumentException(i + " must be > " + x"); – Andreas L. Nov 25 '14 at 19:00
1

That would make it less clear, IMO: 4 must be > 5: what does that mean? But I agree having the values would be helpful: `throw new IllegalArgumentException("i (whose value is " + i + ") must be > x (whose value is " + x + ")");` – JB Nizet Nov 25 '14 at 19:04

Sotirios Delimanolis · Answer 3 · 2014-11-25T18:58:48.197

2

That are no built-in preconditions for method (or constructor) arguments. Either check them at the calling site or within the method (or constructor). You have two options: throw an exception within the method or don't call the method.

edited Nov 25 '14 at 18:58

answered Nov 25 '14 at 18:52

Sotirios Delimanolis

263,859
56
671
702

1

@shekharsuman You parsed it wrong. "Don't call the method" or "throw an exception", parallel to the preceding sentence. – Dave Newton Nov 25 '14 at 18:54
Well, now parallel, but reversed. – Dave Newton Nov 25 '14 at 19:12

How do I restrict the range permitted for the int argument of a method

3 Answers3