Do I need to check null and “” while comparing String in Java?

Question

Suppose I have to compare two String s:

public String checkComaprision(String str) {
    if (!("Hello".equals(str))) {
        System.out.println("String didn't match to hello");
    }

    if ("".equals(str) || null == str || (!str.equals("Hello"))) {
        System.out.println("String didn't match to hello");
    }
}

Here I have used two methods of string comparison insider string. I have read at many places where we compare string with "" and with null before actual comparison. But I think the first case will work properly, and if so, then why should I check for null and blank?

Is there any case where the first comparison will fail? Which is the better approach?

Answer 1

There is no way that "Hello".equals(str)) would ever fail, a string literal simply can never be null . Same thing with "".equals(str) .

The method Objects.equals(Object, Object) automatically checks nulls first.

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To answer your question, using if ("Hello".equals(str)) is just fine, but if you ever want to replace the string literal "Hello" by a variable, you should use Objects.equals(greeting, str) .
Checking for an empty string is absolutely unnecessary.

Answer 2

in your case the firs comparison will always work because we are sure that "hello" is never going to be null. Now if you want have a variable like this

private String s

then you want to compare with another throug a function like yours, you now need to make sure that s is different to null

s!=null

if not, you may have a NullPointerException