Julia 1.0 UndefVarError-变量范围 [英] Julia 1.0 UndefVarError - Scope of Variable

问题描述

我正在从Julia 0.7移到1.0.似乎Julia的变量范围规则从0.7更改为1.0.例如，我想运行一个简单的循环，如下所示:

I am moving from Julia 0.7 to 1.0. It seems that Julia's rule for the scope of variables changed from 0.7 to 1.0. For example, I want to run a simple loop like this:

num = 0
for i = 1:5
    if i == 3
        num = num + 1
    end
end
print(num)

在Julia 0.7(以及大多数其他语言)中，我们可以期望循环后为num = 1.但是，它将在Julia 1.0中产生UndefVarError: num not defined.我知道通过使用let我可以做到这一点

In Julia 0.7 (and in most of other languages), we could expect num = 1 after the loop. However, it will incur UndefVarError: num not defined in Julia 1.0. I know that by using let I can do this

let
num = 0
for i = 1:5
    if i == 3
        num = num + 1
    end
end
print(num)
end

它将打印出1.但是我确实想将num = 1放在循环和let块之外.一些答案建议将所有代码放在let块中，但是在逐行测试时会引起其他问题，包括UndefVarError.有什么方法可以代替使用let阻止吗?谢谢！

It will print out 1. But I do want to get the num = 1 outside the loop and the let block. Some answers suggest putting all code in a let block, but it will incur other problems including UndefVarError while testing line-by-line. Is there any way instead of using let blocking? Thanks!

推荐答案

在此处中对此进行了讨论.

在num变量的循环内如下所示添加global.

Add global as shown below inside the loop for the num variable.

num = 0
for i = 1:5
    if i == 3
        global num = num + 1
    end
end
print(num)

在Julia 1.0.0 REPL中运行:

Running in the Julia 1.0.0 REPL:

julia> num = 0
0
julia> for i = 1:5
           if i == 3
               global num = num + 1
           end
       end
julia> print(num)
1

修改

对于来到Julia的新手，请注意vasja在下面的答案中做出的出色评论:

For anyone coming here new to Julia, the excellent comment made in the answer below by vasja, should be noted:

请记住，在函数内部您不会使用全局函数，因为函数内部的作用域规则符合您的期望:

Just remember that inside a function you won't use global, since the scope rules inside a function are as you would expect:

请参见该答案，这是在同一代码中使用函数而没有范围问题的一个很好的例子.

See that answer for a good example of using a function for the same code without the scoping problem.

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