Ruby:使用Object.send分配变量

Question

有没有办法做这样的事情？

a = Struct.new(:c).new(1)
b = Struct.new(:c).new(2)

a.send(:c)
=> 1

b.send(:c)
=> 2

a.send(:c) = b.send(:c)

最后一行导致错误:

syntax error, unexpected '=', expecting $end
a.send(:c) = b.send(:c)
            ^

Answer 1

a.send(:c=, b.send(:c))

foo.bar = baz不是bar对赋值所遵循的方法的调用- 它是对方法的调用bar=.所以你需要告诉send调用该方法.

Answer 2

如果你事先知道变量名

a.send(:c=, b.send(:c))

如果c是动态变量，那么你可以这样做

c = 'my_key'
a.send("#{c}=", b.send(c))