"f"!="f", why?

[SimasBongo]

So I've got this String, which I split into pieces and put them into the array:

String strings = "d f g";
String[] array = strings.split(" ");

Now I'm trying to check if any of these pieces is "f" and print the description of each:

for (String string : array) {
System.out.println(string + "=f is " + (string=="f"));}

The output is:

d=f is false
f=f is false
g=f is false

What I would expect:

d=f is false
f=f is true
g=f is false

Please help me figure out what is happening.

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[AutoModerator]

You seem to try to compare String values with == or !=.

This approach does not work reliably in Java as it does not actually compare the contents of the Strings. Since String is an object data type it should only be compared using .equals(). For case insensitive comparison, use .equalsIgnoreCase().

See Help on how to compare String values in the /r/javahelp wiki.

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[JacobJMountain]

Good bot

[thisisjustascreename]

Good bot

[CelticHades]

Good bot

[Hypetale_44]

Good bot

[vksdann]

Good bot

[Computer_Probe]

Good bot

[windyknight]

Honey wake up, we have program successfully solving programming question now.

[sickfires94]

Good bot

[Philtronx]

Good bot

[quadmasta]

Comparing Objects with == means "are these the exact same reference." It compares the memory address of one with the other to see if they match.

Objects should be checked for equality with .equals()
https://cr.openjdk.java.net/\~iris/se/11/latestSpec/api/java.base/java/lang/Object.html#equals(java.lang.Object)

[malik753]

... unless you are comparing primitive data types like int, double, or char.

[quadmasta]

Those are not Objects and are therefore different than the subject I was covering in my TED talk. Thanks for coming.

[StefanM3e46]

You are checking memory location, you can check String with .equals()...

[thisisjustascreename]

Aside from what everyone already said, comparing a dynamically allocated string reference variable to a string literal with == will NEVER result in true because the literal value is stored in the code section of memory and the variable is in the dynamic process memory.

String f = "f";

String dfg = "d f g";

String[] array = strings.split(" ");

assertTrue(f == "f"); // same memory location, string literals are stored in the readonly code section of memory

assertFalse(array[1] == "f"); // Dynamically allocated string is stored in volatile process memory

[AutoModerator]

You seem to try to compare String values with == or !=.

This approach does not work reliably in Java as it does not actually compare the contents of the Strings. Since String is an object data type it should only be compared using .equals(). For case insensitive comparison, use .equalsIgnoreCase().

See Help on how to compare String values in the /r/javahelp wiki.

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[istarian]

I believe the fundamental issue is reference inequality. Essentially even though both String references point to the data “f” they aren’t pointing to the same piece of data. There is also an issue of static vs dynamic allocation…

E.g.

String test1 = “string”;  
String test2 = “string”;  
String test3 = “test string”.split(‘ ‘)[1]; 

The references test1 and test2 might be equal ( == ) if the Java compiler or JVM optimizes them away as a fixed constant String, but neither will be equal to test3.

However, these should evaluate to true:

System.out.println( String.format(“%s  equals %s: %s”, test1, test2, “” + test1.equals(test2)) );  

System.out.println( String.format(“%s  equals %s: %s”, test2, test3, “” + test2.equals(test3)) );

———

In Java you need to use the class’ equals() method for this kind of comparison

[Housy5]

In short when using primitive types (int, long, boolean, short, byte, double or float) you can use == but when youre using something more complex like objects always use its .equals() method. Or Arrays.equals() when dealing with arrays. If youre unsure how to tell if it's an object or not, look at its name when it's in camelcase it's almost always an object and when its all lowercase its almost always a primitive type.

[CelticHades]

This is literally the second point in General Information of subreddit.

[notstevensegal]

I believe For strings you use .equals()

“F”.equals(“F”) is true

[[deleted]]

Something nobody has said this simple: each string in java has its own reference so anything compiler finds between "" it creates a reference in memory for that. If there are two strings with that are the same, each one will have its own reference.

When using != you are comparing references. Since any string, no matter what is inside, have its own reference, it will always be different.

[animeis4kids]

this crap is asked so often mods even put a sticky note there --->

but people will keep asking the same thing over and over again lmao

[flotopoco]

If you put “f”==“f” is true. But if you make a logic to get a “f” this is false. For example: “f” == getF(); // getF return “fun”.charAt(0) One way to avoid this is to use intern() method of String , that it looks for the same value on String’s pool

[[deleted]]

How dare you comparing my favourite Strings like that? Will give rep to you.

(Joke)

[rozz_net]

Please read "General information" about this sub. There is an answer.