使用 pry-byebug 设置条件断点

Set conditional breakpoint with pry-byebug

给定以下代码:

require 'pry-byebug'

10.times.with_index do |i|
  binding.pry
  puts i
end

我想在里面 "loop until i == 5 then break" pry-byebug。从 运行 到 help break 看来你可以通过 "breakpoint #" 或 "line #" 来识别断点。您似乎也可以使用条件,但我没有让它工作:

正在尝试在行 #:

上设置断点
$  ruby foo.rb

From: /Users/max/Dropbox/work/tmp/pry_debug/foo.rb @ line 5 :

    1: require 'pry-byebug'
    2:
    3: 10.times.with_index do |i|
    4:   binding.pry
 => 5:   puts i
    6: end

[1] pry(main)> break foo.rb:5 if i == 5

  Breakpoint 1: /Users/max/Dropbox/work/tmp/pry_debug/foo.rb @ 5 (Enabled) Condition: i == 5


    2:
    3: 10.times.with_index do |i|
    4:   binding.pry
 => 5:   puts i
    6: end


[2] pry(main)> continue
0

From: /Users/max/Dropbox/work/tmp/pry_debug/foo.rb @ line 5 :

    1: require 'pry-byebug'
    2:
    3: 10.times.with_index do |i|
    4:   binding.pry
 => 5:   puts i
    6: end

[2] pry(main)> i
=> 1

先设置断点再把条件放在断点上:

➜  ~/D/w/t/pry_debug  ruby foo.rb

From: /Users/max/Dropbox/work/tmp/pry_debug/foo.rb @ line 5 :

    1: require 'pry-byebug'
    2:
    3: 10.times.with_index do |i|
    4:   binding.pry
 => 5:   puts i
    6: end

[1] pry(main)> break foo.rb:15

  Breakpoint 1: /Users/max/Dropbox/work/tmp/pry_debug/foo.rb @ 15 (Enabled)



[2] pry(main)> break --condition 1 i == 5
[3] pry(main)> c
0

From: /Users/max/Dropbox/work/tmp/pry_debug/foo.rb @ line 5 :

    1: require 'pry-byebug'
    2:
    3: 10.times.with_index do |i|
    4:   binding.pry
 => 5:   puts i
    6: end

[3] pry(main)> i
=> 1

如您所见,在这两种情况下 pry-byebug 都不符合条件,因为它停止得太早了。我如何让它工作?

我认为一种简单的方法是在您的代码中执行此操作:

require 'pry-byebug'

10.times.with_index do |i|
  binding.pry if i == 5
  puts i
end

或者如果你想在异常时停止,你可以这样做:

require 'pry-byebug'

10.times.with_index do |i|
  puts i rescue binding.pry # if the puts command throws an exception, you'll be taken to pry
end

从循环内部删除 binding.pry 并将其放在 10.times:

之前
binding.pry
10.times.with_index do |i|

然后运行代码。当它到达断点时,然后使用您想要的条件和 continue.

设置一个新断点

回到你的实际代码。不要这样做:

10.times.with_index do |i|

相反,这将做几乎相同的事情,但它更简单:

10.times do |i|

这是 Ruby 正在做的事情:

>> 2.times.with_index.to_a
[
  [0] [
    [0] 0,
    [1] 0
  ],
  [1] [
    [0] 1,
    [1] 1
  ]
]

对比

>> 2.times.to_a
[
  [0] 0,
  [1] 1
]

首先是传入 [0,0][1,1] 等数组,因此为了正确起见,您的块参数需要类似于:

10.times.with_index do |i, j|

第二个只传递当前 "times" 值,导致代码更简单。