Getting ArrayIndexOutOfBoundsException Exceptions

问题

Past few days Getting ArrayIndexOutOfBoundsException .I know this type of questions already asked on #SO . I tried .

java.lang.ArrayIndexOutOfBoundsException: length=1; index=1

Code

String[] Child_DOB = "KUSHAGRA (SON)-07/05/94AANVI (DAUGHTER)-12/06/00 VARENYA (SON) - 26/12/05";
        ArrayList<String> children_List = new ArrayList<String>();
        ArrayList<Integer> Length_List = new ArrayList<Integer>();
        String Children_Details_str = "";

        int i = 0;
        while (i < Child_DOB.length) {
            String name_dob = Child_DOB[i] + "  " + Child_DOB[i + 1];//this line
            if (i > 3)
                Children_Details_str = Children_Details_str + "\n" + name_dob;
            else
                Children_Details_str = Children_Details_str + name_dob + "  ";
            children_List.add(name_dob);
            Length_List.add(Child_DOB[i].length());
            Length_List.add(Child_DOB[i + 1].length());
            i = i + 2;
        }

May I know what is the correct way to achieve my objective? Any help would be greatly appreciated

回答1:

This is because of line as you pointed out:

String name_dob = Child_DOB[i] + "  " + Child_DOB[i + 1];//this line

Say you have only one element in your Child_DOB array, you would enter your while loop, and with i = 0 try to access i+1 i.e. 1st element. Array starts with 0th index and hence accessing element at index 1 would throw ArrayIndexOutOfBoundsException. To avoid this one option would be:

while (i < Child_DOB.length - 1) {//go until second last element

This would work if your array has even number of elements. If you have odd number of elements then you would miss the last name (which should be fine as per your logic.)

if you really dont want to miss the last element, you could do something like:

if ((Child_DOB.length & 0X1) == 1) {
    Length_List.add(Child_DOB[Child_DOB.length - 1].length());
    //append to Children_Details_str.. better use StringBuilder here
}

回答2:

Change your while loop condition as:

 while (i < Child_DOB.length - 1)

Explanation:

For example, Child_DOB.length is 5 and i value is 4, In your code:

    int i = 4;
    while (i < 5) {
        String name_dob = Child_DOB[4] + "  " + Child_DOB[4 + 1];

Here Child_DOB[5] causes ArrayIndexOutOfBoundsException because array index start from 0 and your Child_DOB index's are [0 1 2 3 4].

Hope this will help~

回答3:

You have only one element in Child_DOB so Child_DOB.length=1, and inside while loop you are trying to access 2nd element Child_DOB[i + 1] in first iteration itself... So you are getting exception... Can you tell exactly what output you want??

回答4:

Use-

while (i < Child_DOB.length - 1)

Instead of -

while (i < Child_DOB.length)

回答5:

There is only one String in CHILD_DOB. In your while statement you typed Child_DOB[i + 1]; which accesses CHILD_DOB[1] but there is only one item in CHILD_DOB which is CHILD_DOB[0] Try this

String[] Child_DOB = new String[] {"KUSHAGRA",
 "(SON)-07/05/94AANVI", "(DAUGHTER)-12/06/00", "VARENYA", "(SON)", "-", "26/12/05"};

Which I think would fix your code.

来源：https://stackoverflow.com/questions/43511894/getting-arrayindexoutofboundsexception-exceptions

标签

android

indexoutofboundsexception