Ruby:
true == true == true
syntax error, unexpected tEQ
vs. JavaScript:
true == true == true
// => true
vs. C:
1 == 1 == 1
// => 1
If the two operands are not of the same type, JavaScript converts the operands then applies strict comparison. If either operand is a number or a boolean, the operands are converted to numbers if possible; else if either operand is a string, the other operand is converted to a string if possible.
If the left side of the expression is "falsey", the expression will return the left side. If the left side of the expression is "truthy", the expression will return the right side. That's it. So in false && false , the left side is "falsey", so the expression returns the left side, false .
The && requires that both items are true – think of it as having two statements on either side (rather than just true or false itself). The && means, essentially: “Are BOTH of these statements true?” If the answer is yes, then it returns true. in any other case, it returns false.
Association direction, which controls the order of operators having their arguments evaluated, is not defined for the ==
method, same as for ===
, !=
, =~
and <=>
methods as well (all of which have the same precedence and form a separate precedence group exclusively).
Documentation
Thus evaluation order in case of multiple operators from the list mentioned above being chained in a row should be set explicitly via either
parenthesis ()
:
(true == true) == true # => true
true == (true == true) # => true
or dot operator .
(can be omitted for the last equality check in a row):
true .== true == true # => true
If I understand the question correctly value_a == value_b == value_c
should only return true if they are all equal using == as the comparison operater as shown in this method
# version 1
def compare_3_values(a, b, c)
a == b && a == c && b == c
end
there is another possible expected outcome though. to implement this as shown in the previous answer:
#version 2
def compare_3_values(a, b, c)
(a == b) == c
end
The results are worlds apart.
JavaScript always uses version 2 which is pretty useless as the 3rd item is always being compared against true or false (0 or 1 if the 3rd item is an integer) that's why false == false == true
returns true.
The good news is that because ruby gives a syntax error it's the only language that can implement this without breaking everyone's code.
for any other language it would break so much code that even if it were implemented in a later major version there would need to be a flag/setting to turn this on or off for years to come, hence it will never be worthwhile.
Some interesting results in Ruby
false .== false == true
=> true
false .== true == false
=> true
true .== false == false
=> true
false .== false == false
=> false
true .== true == false
false
And in javascript
false == false == true
=> true
false == true == false
=> true
true == false == false
=> true
false == false == false
=> false
true == true == false
=> false
Edit tested in C as well, acts similar to JavaScript in that it compares the result of the first two values against the third value
The first answer is excellent, but just in case it's not completely clear (and people asking why), here are few more examples.
In C, the ==
operator is left-to-right associative and boolean is represented as 1 (true) and 0 (false), so the first 1 == 1
evaluates to 1
(true) and then you are evaluating the result of first expression with the second. You can try:
2 == 2 == 2 // => 0
Which in C, is evaluated as:
(2 == 2) == 2
1 == 2 // => 0
In Javascript, similarly to C, ==
is left to right associative. Let's try with 0 this time (although the same example from C would work as well):
0 == 0 == 0
false
Again:
0 == 0 == 0
true == 0 // => false
In Ruby ==
does not have associative properties, ie. it can't be used multiple times in single expression, so that expression can't be evaluated. Why that decision was made is a question for the author of the language. Further, Ruby doesn't define numeric 1 as a boolean, so 1 == true
evaluates to false.
The second answer states there are some "weird" cases in Ruby, but they all evaluate as expected:
(1 == 1) == 1
true == 1 # => false
1 == (1 == 1)
1 == true # => false
1 .== 1 == 1
(1 == 1) == 1
true == 1 # => false
false .== false == true
(false == false) == true
true == true # => true
false .== true == false
(false == true) == false
false == false # => true
true .== false == false
(true == false) == false
false == false # => true
false .== false == false
(false == false) == false
true == false # => false
true .== true == false
(true == true) == false
true == false # => false
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