調試/檢查功能

Question

考慮以下人爲的例子內的值：

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 

我想檢查的result1值，result2，和result3由時間我到達線6然而，當我通過步驟代碼，在任何時候，做任何這些「計算」的出現可供查閱：

*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 

也就是說我們有像result1 :: a = _線，但沒有像result1 :: a = 2。這是因爲哈斯克爾懶惰的本質嗎？如果是的話，是否有辦法檢查result1,result2和result3當時值是綁定到他們？謝謝

Answer 1

你是正確的行爲是由於懶惰。要強制進行評估，您可以執行：

[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 = _