Haskell在递归时打印？ [英] Haskell Print while in recursion?

问题描述

foo:: Int -> Int -> Int
foo z x = if (z < 100) 
             then z * foo (z+(x*z)) z
             else z

每次从它本身调用它时，你会如何输出（整数z）输出？你可以有返回IO和Int的函数吗？你需要一个辅助函数吗？

How would you print out(the integer z) an output every time it gets called from itself?Can you have function that returns an IO and Int? Do you need a secondary function?

推荐答案

为了简单起见，你可以使用 trace 。然而，它并不欢迎真正的生产代码，因为它会破坏参考透明度。
trace 需要打印一个字符串并返回一个值。

For simplicity, you can use trace. However it's not welcomed for real production code as it breaks referential transparency. trace takes a String to print and a value to return.

import Debug.Trace

foo:: Int -> Int -> Int
foo z x = trace ("z = " ++ show z) $ if (z < 100) 
    then z * foo (z+(x*z)) z
    else z

*Main> foo 1 2
 z = 1
 z = 3
 z = 6
 z = 24
 z = 168
 72576

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