julia - 带有嵌套在 for 循环中的 if block 的 Julia 变量的范围

Question

for i=1:5
    if(i==1)
        z = i
    end
    println("i = $i, z = $z")
end

i = 1, z = 1
ERROR: UndefVarError: z not defined
Stacktrace:
 [1] top-level scope at ./REPL[6]:5 [inlined]
 [2] top-level scope at ./none:0

上述脚本的行为令我感到困惑。有人能帮我理解为什么在 i=1 时打印工作但在 i=2 时失败。

Answer 1

@Wookies-Will-Code 的答案是正确的，原因如下。

正如 Julia 手册解释的那样 here :

for loops, while loops, and Comprehensions have the following behavior: any new variables introduced in their body scopes are freshly allocated for each loop iteration

这意味着在每次迭代结束时 z被遗忘了。

为了确定 z的值在迭代中持续(以及在 for 循环结束之后)定义 z在循环之前。例如像这样:

function mytest()
    local z
    for i=1:5
        if i==1
            z = i
        end
        println("i = $i, z = $z")
    end
end

现在你有:

julia> mytest()
i = 1, z = 1
i = 2, z = 1
i = 3, z = 1
i = 4, z = 1
i = 5, z = 1

请注意，情况在全局范围内有所不同。即使 z是全局定义的，默认情况下不重用:

julia> z = 100
100

julia> for i=1:5
           if i==1
               z = i
           end
           println("i = $i, z = $z")
       end
i = 1, z = 1
ERROR: UndefVarError: z not defined
Stacktrace:
 [1] top-level scope at .\REPL[7]:5 [inlined]
 [2] top-level scope at .\none:0

原因是Julia对待 z作为 for 循环的局部，因为 z被分配到循环内部。这是一个编译时特性检查，所以即使 if条件总是失败你会得到这个错误:

julia> z = 100
100

julia> for i=1:5
           if false
               z = i
           end
           println("i = $i, z = $z")
       end
ERROR: UndefVarError: z not defined
Stacktrace:
 [1] top-level scope at .\REPL[18]:5 [inlined]
 [2] top-level scope at .\none:0

您必须分配给 z使用 global关键字使其工作:

julia> z = 100
100

julia> for i=1:5
           if i==1
               global z = i
           end
           println("i = $i, z = $z")
       end
i = 1, z = 1
i = 2, z = 1
i = 3, z = 1
i = 4, z = 1
i = 5, z = 1