为什么不'for'在'go'里面工作？

Question

为什么调用此函数不会打印任何内容？

(defn test-go-loop []
  (go (for [a (cycle [:a :b :c])]
        (do (println a) (<! (timeout 1000))))))

Answer 1

for懒惰评估,你的代码中没有任何东西要求结果for.试试doseq:

(defn test-go-loop []
  (go (doseq [a (cycle [:a :b :c])]
        (println a)
        (<! (timeout 1000)))))