Debugging 调试/检查函数内的值

考虑以下人为的例子：

module Main where
myadd3 first second third = 
  let result1 = first               -- line 3
      result2 = second              -- line 4
      result3 = third               -- line 5
  in result1 + result2 + result3    -- line 6

我想在到达第6行时检查

result1

、

result2

和

result3

的值。然而，当我一步一步地浏览代码时，这些“计算”中的任何一个似乎都不可供检查：

*Main> :break 3
Breakpoint 0 activated at Main4.hs:3:17-21
*Main> :break 4
Breakpoint 1 activated at Main4.hs:4:17-22
*Main> :break 5
Breakpoint 2 activated at Main4.hs:5:17-21
*Main> :break 6
Breakpoint 3 activated at Main4.hs:6:6-32
*Main> myadd3 2 3 4
Stopped at Main4.hs:6:6-32
_result :: a = _
result1 :: a = _
result2 :: a = _
result3 :: a = _
[Main4.hs:6:6-32] *Main> :step
Stopped at Main4.hs:6:6-22
_result :: a = _
result1 :: a = _
result2 :: a = _
[Main4.hs:6:6-22] *Main> :step
Stopped at Main4.hs:3:17-21
_result :: Integer = _
first :: Integer = 2
[Main4.hs:3:17-21] *Main> :step
Stopped at Main4.hs:4:17-22
_result :: Integer = _
second :: Integer = 3
[Main4.hs:4:17-22] *Main> :step
Stopped at Main4.hs:5:17-21
_result :: Integer = _
third :: Integer = 4
[Main4.hs:5:17-21] *Main> :step
9

---

也就是说，我们有类似于

result1:：a=\ucode>的行，但没有类似于
result1:：a=2的行。这是因为haskell的懒惰本性吗？如果是，那么在值绑定到它们的时候，有没有办法检查
result1
、
result2
、和
result3
？谢谢
你是对的，这种行为是由于懒惰造成的。要强制评估，您可以执行以下操作：

[debug.hs:6:6-32] *Main> :print result1
result1 = (_t1::a)
[debug.hs:6:6-32] *Main> :force result1
*** Ignoring breakpoint
result1 = 2
[debug.hs:6:6-32] *Main> :step
Stopped at debug.hs:6:6-22
_result :: Integer = _
result1 :: Integer = 2
result2 :: Integer = _