how IllegalArgumentExcpetion works in java

Question

I'd like to use IllegalArgumentExcpetion in my script but I can't find any information anywhere. I show what i already have:

public int getId() {
        return id;
    }

    public void setId(int id) {

        if(id<=0) {
            throw new IllegalArgumentException("XX must be a positive integer greater than 0!!");           
        }
        this.m2=m2;
    }

I'd like the IllegalArgumentException show the message “XX must be a positive integer greater than 0!!” and XX means id. My doubt is that I don't know how I can transform the XX into the id

Concat that : `throw new IllegalArgumentException( id + " must be a positive integer greater than 0!!");` — davidxxx, Dec 15 '18 at 17:38
By the way, `this.m2=m2;` does nothing useful: you are setting a field to itself. Perhaps you mean `this.m2=id;`? — Andy Turner, Dec 15 '18 at 17:54

score 2 · Accepted Answer · answered Dec 15 '18 at 17:41

There's nothing special about IllegalArgumentException. You just need to construct the message, e.g. by string concatenation and pass it to the constructor:

throw new IllegalArgumentException(id + " must be a positive integer greater than 0!!");

how IllegalArgumentExcpetion works in java

1 Answers1