146 lines
2.2 KiB
CoffeeScript
146 lines
2.2 KiB
CoffeeScript
# CoffeeScript's operations should be chainable, like Python's.
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ok 500 > 50 > 5 > -5
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ok true is not false is true is not false
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ok 0 is 0 isnt 50 is 50
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ok 10 < 20 > 10
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ok 50 > 10 > 5 is parseInt('5', 10)
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# Make sure that each argument is only evaluated once, even if used
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# more than once.
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i = 0
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func = -> i++
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ok 1 > func() < 1
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# `==` and `is` should be interchangeable.
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a = b = 1
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ok a is 1 and b is 1
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ok a == b
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ok a is b
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# Ensure that chained operations don't cause functions to be evaluated more
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# than once.
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val = 0
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func = -> val = + 1
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ok 2 > (func null) < 2
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ok val is 1
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# Allow "if x not in y"
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obj = {a: true}
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ok 'a' of obj
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ok 'b' not of obj
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# And for "a in b" with array presence.
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ok 100 in [100, 200, 300]
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array = [100, 200, 300]
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ok 100 in array
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ok 1 not in array
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list = [1, 2, 7]
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result = if list[2] in [7, 10] then 100 else -1
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ok result is 100
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# And with array presence on an instance variable.
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obj = {
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list: [1, 2, 3, 4, 5]
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in_list: (value) -> value in @list
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}
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ok obj.in_list 4
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ok not obj.in_list 0
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# Non-spaced values still work.
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x = 10
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y = -5
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ok x*-y is 50
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ok x*+y is -50
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# Compound operators.
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one = two = null
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one or= 1
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two or= 2
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ok one is 1
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ok two is 2
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one and= 'one'
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two and= 'two'
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ok one is 'one'
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ok two is 'two'
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# Compound assignment should be careful about caching variables.
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list = [0, null, 5, 10]
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count = 1
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key = ->
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count += 1
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list[key()] or= 100
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ok list.join(' ') is '0 5 10'
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count = 0
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list[key()] ?= 100
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ok list.join(' ') is '0 100 5 10'
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# Ensure that RHS is treated as a group.
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a = b = false
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a and= b or true
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ok a is false
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# Bitwise operators:
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ok (10 & 3) is 2
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ok (10 | 3) is 11
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ok (10 ^ 3) is 9
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ok (10 << 3) is 80
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ok (10 >> 3) is 1
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ok (10 >>> 3) is 1
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num = 10; ok (num <<= 3) is 80
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num = 10; ok (num >>= 3) is 1
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num = 10; ok (num >>>= 3) is 1
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num = 10; ok (num &= 3) is 2
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num = 10; ok (num ^= 3) is 9
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num = 10; ok (num |= 3) is 11
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# Compound assignment with implicit objects.
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obj = undefined
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obj ?=
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one: 1
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ok obj.one is 1
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obj and=
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two: 2
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ok not obj.one
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ok obj.two is 2
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# Compound assignment as a sub expression.
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[a, b, c] = [1, 2, 3]
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ok (a + b += c) is 6
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ok a is 1
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ok b is 5
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ok c is 3
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# Instanceof.
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ok new String instanceof String
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ok new Number not instanceof String
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