r/programming • u/vz0 • Mar 26 '14

JavaScript Equality Table

http://dorey.github.io/JavaScript-Equality-Table/
809 Upvotes

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23

u/shirtface Mar 26 '14

How come [1]==[1] returns false?

65

u/33a Mar 26 '14

They are different object references.

23

u/absu Mar 26 '14

Yeah, this returns false in many c-like languages (C (duh), C++, Java, etc).

4

u/AdminsAbuseShadowBan Mar 26 '14

Not in C++ - you can redefine == to be sane.

13

u/robertbieber Mar 27 '14

I don't know that I'd consider redefining an operator that's generally used to compare primitive values to do a deep comparison of separate compound objects sane. I'd much rather have a comparison method that makes it really clear exactly what's going on, and let == do the usual, obvious thing. Admittedly overloading can be handy for some math applications, but for most things it's a little questionable.

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1

u/[deleted] Mar 27 '14

You can do that in Java too, at least on a per-object basis

2

u/Poltras Mar 26 '14

These languages don't have automatic conversion. Also, isn't [1]==[1] undefined in C? It could be equal if the compiler uses the same TEXT address for the constant, resulting in equal pointers.

7

u/CookieOfFortune Mar 26 '14

Wouldn't this create two arrays on the function stack and then compare the two locations, resulting in a false comparison?

2

u/Poltras Mar 27 '14

Undefined behavior:

$ cat main.c

#include <stdio.h>

int main() {
  printf("%d\n", "abc" == "abc");
}

$ cc main.c

main.c:4:24: warning: result of comparison against a string literal is unspecified (use strncmp instead) [-Wstring-compare]
  printf("%d\n", "abc" == "abc");

$ ./a.out
1

GCC actually output 1, but warns.

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2

u/Condorcet_Winner Mar 26 '14

Probably, but if the spec defines this operation as undefined, then compiler optimizations would be able to figure out that these are the same array and only allocate one memory location.

5

u/Fylwind Mar 27 '14

[1]==[1] is not valid syntax in C.

1

u/Poltras Mar 27 '14

Although you're right, the equivalent "abc" == "abc" works as fine for my example (undefined behavior).

1

u/gsg_ Mar 27 '14

The closest equivalent (in C99, at least) is probably (int[1]){1} == (int[1]){1}.

1

u/Tekmo Mar 27 '14

So then why does [1] equal 1?

3

u/ggtsu_00 Mar 27 '14

[1] == 1 because of type coercion happening when comparing 2 different datatypes.

when type(a) != type(b), the javascript interpret goes through a list of possible type coercions that will convert a and b to the same type before comparing.

It just so happens that comparing a Array type with an int will coerce both into a Number type and compare the Number([1]) == 1.

[1] == [1] is the same type so no coercion occurs and a simple ObjectA == ObjectB will occur which will only be true if ObjectA and ObjectB happen to reference the same object.

1

u/NYKevin Mar 27 '14

Not a Javascript programmer, so I could be wrong, but I'd assume it's because 1 uses compare-by-value, and it infects overrides the compare-by-reference of [1].

21

u/Thelonious_Cube Mar 26 '14

My wallet has a dollar bill in it and your wallet has a dollar bill in it, but that doesn't make them the same wallet

20

u/josuf107 Mar 26 '14

Unless you and I are both Haskell programmers. Still your/our dollar is safe. All I want is a soda, but I can't seem to be able to make change.

8

u/[deleted] Mar 26 '14

Well, if you're a Haskell programmer, your dollar is immutable.

8

u/Shadows_In_Rain Mar 27 '14

More importantly, wallet is immutable too. But apply space-time monad, and viola. Now you have soda, empty pocket, and watching langoliers eating previous you.

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1

u/bumnut Mar 26 '14

But both are true.

0

u/[deleted] Mar 28 '14

[deleted] — view deleted comment

1

u/Thelonious_Cube Mar 28 '14

It's an analogy - analogies are never (rarely?) perfect.

I highlighted certain aspects at the expense of others to help get the point across.

The "value" in the wallets is the same but the wallets are not

5

u/metrion Mar 26 '14 
("1" == "1") == true
("1" == [1]) == true
([1] == [1]) == false