This is a Java string problem. I use the substring(beginindex) to obtain a substring.
Considering String s="hello", the length of this string is 5. However when I use s.substring(5) or s.substring(5,5) the compiler didn't give me an error. The index of the string should be from 0 to length-1.
Why it doesn't apply to my case? I think that s.substring(5) should give me an error but it doesn't.
3 Answers
Because the endIndex is exclusive, as specified in the documentation.
IndexOutOfBoundsException - if the beginIndex is negative, or endIndex is larger than the length of this String object, or beginIndex is larger than endIndex.
I think when I use s.substring(5), it should give me error while it didn't
Why would it be?
Returns a new string that is a substring of this string. The substring begins with the character at the specified index and extends to the end of this string.
Since the beginIndex is not larger than the endIndex (5 in your case), it's perfectly valid. You will just get an empty String.
If you look at the source code:
1915 public String substring(int beginIndex) {
1916 return substring(beginIndex, count);
1917 }
....
1941 public String substring(int beginIndex, int endIndex) {
1942 if (beginIndex < 0) {
1943 throw new StringIndexOutOfBoundsException(beginIndex);
1944 }
1945 if (endIndex > count) {
1946 throw new StringIndexOutOfBoundsException(endIndex);
1947 }
1948 if (beginIndex > endIndex) {
1949 throw new StringIndexOutOfBoundsException(endIndex - beginIndex);
1950 }
1951 return ((beginIndex == 0) && (endIndex == count)) ? this :
1952 new String(offset + beginIndex, endIndex - beginIndex, value);
1953 }
Thus s.substring(5); is equivalent to s.substring(5, s.length()); which is s.substring(5,5); in your case.
When you're calling s.substring(5,5);, it returns an empty String since you're calling the constructor(which is private package) with a count value of 0 (count represents the number of characters in the String):
644 String(int offset, int count, char value[]) {
645 this.value = value;
646 this.offset = offset;
647 this.count = count;
648 }
2 Comments
user3382017
but it didn't have the index ==5. but I can use s.substring(5)?why
Alexis C.
@user3382017
s.substring(5); is equivalent to s.substring(5, s.length()); which is s.substring(5,5); for "Hello".Because substring is defined as such, you can find it in the Javadoc of String.substring
@exception IndexOutOfBoundsException if the beginIndex is negative, or endIndex is larger than the length of this String object, or beginIndex is larger than endIndex.
It's useful in many cases that you can always create a substring that starts after a character in a String.
Because endIndex can be the length of the string, and beginIndex can be as large as endIndex (but not larger), it is also okay for beginIndex to be equal to the length of the string.
1 Comment
aliteralmind
+1. You just taught me something. minIndex can be equal to length. I never did it, so I assumed it was illegal.
In first case (s.substring(5)), Oracle docs says
...
IndexOutOfBoundsException - if beginIndex is negative or larger than the length of this String object.
...
In second case (s.substring(5,5)), it says that
...
IndexOutOfBoundsException - if the beginIndex is negative, or endIndex is larger than the length of this String object, or beginIndex is larger than endIndex
...
s.substring(s.length())is silly but valid