Julia 0.3中Docile.jl的示例用法
我是朱莉娅的新手。我对使用添加文档到现有项目感兴趣。根据,Docstring已经添加到Julia 0.4中,但我想让Docstring在0.3中工作 不过,上面文章中的示例在0.3中不起作用<代码>帮助(f)不显示文档字符串,Julia 0.3中Docile.jl的示例用法,julia,documentation-generation,Julia,Documentation Generation,我是朱莉娅的新手。我对使用添加文档到现有项目感兴趣。根据,Docstring已经添加到Julia 0.4中,但我想让Docstring在0.3中工作 不过,上面文章中的示例在0.3中不起作用帮助(f)不显示文档字符串,@doc f返回错误。有人能帮我看看少了什么吗 $ julia _ _ _ _(_)_ | A fresh approach to technical computing (_) | (_) (_) |
@doc f
返回错误。有人能帮我看看少了什么吗
$ julia
_
_ _ _(_)_ | A fresh approach to technical computing
(_) | (_) (_) | Documentation: http://docs.julialang.org
_ _ _| |_ __ _ | Type "help()" for help.
| | | | | | |/ _` | |
| | |_| | | | (_| | | Version 0.3.6 (2015-01-08 22:33 UTC)
_/ |\__'_|_|_|\__'_| | Official http://julialang.org release
|__/ | x86_64-linux-gnu
julia> using Docile
julia> @doc """
Compute 2 times x minus y squared.
""" ->
function f(x::Float64, y::Float64)
return 2x - y^2
end
f (generic function with 1 method)
julia> help(f)
INFO: Loading help data...
f (generic function with 1 method)
julia> @doc f
ERROR: @doc: use `->` to separate docs/object:
(:f,)
您可能希望结合使用Docile.jl。下面的会话显示,如果您只记录函数而不使用Lexicon,则任何函数的帮助文本都不会更改。帮助文本仅在您使用Lexicon键入
后反映您的文档。请注意,在下面,我是通过在repl中键入一个?
来获取帮助文本的,而不是通过使用函数help
_ _ _(_)_ | A fresh approach to technical computing
(_) | (_) (_) | Documentation: http://docs.julialang.org
_ _ _| |_ __ _ | Type "help()" for help.
| | | | | | |/ _` | |
| | |_| | | | (_| | | Version 0.3.6 (2015-02-17 22:12 UTC)
_/ |\__'_|_|_|\__'_| | Official http://julialang.org/ release
|__/ | x86_64-unknown-linux-gnu
julia> using Docile
julia>
julia> @doc """
Compute 2 times x minus y squared.
""" ->
function f(x::Float64, y::Float64)
return 2x - y^2
end
f (generic function with 1 method)
help?> f
INFO: Loading help data...
f (generic function with 1 method)
julia> using Lexicon
help?> f
f (generic function with 1 method)
INFO: Parsing documentation for Docile.
INFO: Parsing documentation for Lexicon.
INFO: Parsing documentation for Docile.Interface.
[method]
.f(x::Float64, y::Float64)
Compute 2 times x minus y squared.
Details:
source: (3,"none")