In Ruby, how to implement 20 - point and point - 20 using coerce() ?
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by Jian Lin
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Published on 2010-05-10T08:19:08Z
Indexed on
2010/05/10
8:24 UTC
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In Ruby, the operation of
point - 20
20 - point
are to be implemented.
But the following code:
class Point
attr_accessor :x, :y
def initialize(x,y)
@x, @y = x, y
end
def -(q)
if (q.is_a? Fixnum)
return Point.new(@x - q, @y - q)
end
Point.new(@x - q.x, @y - q.y)
end
def -@
Point.new(-@x, -@y)
end
def *(c)
Point.new(@x * c, @y * c)
end
def coerce(something)
[self, something]
end
end
p = Point.new(100,100)
q = Point.new(80,80)
p (-p)
p p - q
p q - p
p p * 3
p 5 * p
p p - 30
p 30 - p
Output:
#<Point:0x2424e54 @x=-100, @y=-100>
#<Point:0x2424dc8 @x=20, @y=20>
#<Point:0x2424d3c @x=-20, @y=-20>
#<Point:0x2424cc4 @x=300, @y=300>
#<Point:0x2424c38 @x=500, @y=500>
#<Point:0x2424bc0 @x=70, @y=70>
#<Point:0x2424b20 @x=70, @y=70>
30 - p
will actually be taken as p - 30
by the coerce function. Can it be made to work?
I am actually surprise that the -
method won't coerce the argument this way:
class Fixnum
def -(something)
if (/* something is unknown class */)
a, b = something.coerce(self)
return -(a - b) # because we are doing a - b but we wanted b - a, so it is negated
end
end
end
that is, the function returns a negated version of a - b
instead of just returning a - b
.
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