例如:
for(int i = 0; i < 100; i += 2)
如果2%=0,我可以同时使用 lo或检查,但我很想知道,是否没有办法处理 lo。 我也可以使用<代码>,用于(int i = 0; i <50; i++)和乘以2,但两者似乎都是正确的。 我很想知道,这样做是否简单、直截了当。
(0...100).step(2) do |i|
# loop body
end
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