朱莉娅的lpad函数和字符串插值的奇怪行为

Question

以下代码使我感到惊讶：

julia> for person in 1:12
           println("person is $(lpad(2, person))")
       end
person is 2
person is  2
person is   2
person is    2
person is     2
person is      2
person is       2
person is        2
person is         2
person is          2
person is           2
person is            2

直到我意识到我以相反的顺序输入lpad函数的参数之前，这似乎是完全神秘的。修复它：

julia> for person in 1:12
           println("person is $(lpad(person, 2))")
        end
person is  1
person is  2
person is  3
person is  4
person is  5
person is  6
person is  7
person is  8
person is  9
person is 10
person is 11
person is 12

但是，肯定发生了一些我无法理解的lpad和意外的输出。

下面是docs中lpad函数的签名：

lpad(s, n::Integer, p::Union{AbstractChar,AbstractString}=' ') -> String

有什么建议吗？

Answer 1

由于某种原因，您的循环计数器称为“人”。因此它从1循环到12。

从函数docs中，第二个参数确定 打印的第一个参数：

lpad(s, n::Integer, p::Union{AbstractChar,AbstractString}=' ') -> String

Stringify s and pad the resulting string on the left with p to make
it n characters (code points) long. If s is already n characters long,
an equal string is returned. Pad with spaces by default.

所以您的循环计数器是个人，这也是空格数 在第一个版本的输出的每一行的'2'之前打印。