Java String Comparison Issues Explained

javastring

String[] letters  = {"A", "B", "C", "D", "E", "F", "G", "H", "I", "L"};

Scanner inp = new Scanner(System.in);
String input = (inp.nextLine());
String[] cord = input.split("");

for(int x = 0; x < 10; x++)
    if(letters[x] == cord[1])
        System.out.println("Fk yeah!");

Why the Fk yeah! never happens if I input one of A-L letters?

Best Answer

Strings are objects. The == compares objects by reference, not by their internal value.

There are 2 solutions:

Use String#equals() method instead to compare the value of two String objects.
```
if (letters[x].equals(cord[1]))
```

Use char instead of String. It's a primitive, so == will work.

char[] letters  = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'L'};

Scanner inp = new Scanner(System.in);
String input = (inp.nextLine());
char[] cord = input.toCharArray();

for (int x = 0; x < 10; x++)
    if (letters[x] == cord[1])
        System.out.println("Fk yeah!");

Related Solutions

Java String Comparison – Are Both Strings Equal?

s2 should point to same "learnJava" as its already present in StringConstantPool.

Nope - the string pool is only used for constant strings unless you call intern.

So the string pool contains "learn", "Java" and "learnJava"... but s1 refers to a string not in the string pool, because the concatenation is performed at execution time.

If you had a compile-time constant expression, it would be different:

String s1 = "learn" + "Java"; // Constant expression
String s2 = "learnJava";
System.out.println(s1 == s2); // true, guaranteed

Or if you want to keep your execution-time concatenation:

String s1 = "learn";
s1 += "Java";
s1 = s1.intern();
String s2 = "learnJava";
System.out.println(s1 == s2); // true

In general though, just don't compare strings with ==... it's almost always a bad idea.

Best Answer

Related Solutions

Java String Comparison – Are Both Strings Equal?

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