Ruby变量(数组)赋值误解(使用push方法)

时间:2010-09-08 22:32:52

标签: ruby

我发现我对Ruby或编程理论或两者的理解存在缺陷。看看这个代码:

#!/usr/bin/ruby -w
@instance_ar = [1,2,3,4]
local_ar = @instance_ar
local_ar_2 = local_ar
###
irrelevant_local_ar = [5,6,7,8]
###
for i in irrelevant_local_ar
    local_ar_2.push(i)
end
count = 0
for i in local_ar_2
    puts "local_ar_2 value: #{i} and local_ar value: #{local_ar[count]} and @instance_ar value: #{@instance_ar[count]}\n"
    count += 1
end

输出是

local_ar_2 value: 1 and local_ar value: 1 and @instance_ar value: 1
local_ar_2 value: 2 and local_ar value: 2 and @instance_ar value: 2
local_ar_2 value: 3 and local_ar value: 3 and @instance_ar value: 3
local_ar_2 value: 4 and local_ar value: 4 and @instance_ar value: 4
local_ar_2 value: 5 and local_ar value: 5 and @instance_ar value: 5
local_ar_2 value: 6 and local_ar value: 6 and @instance_ar value: 6
local_ar_2 value: 7 and local_ar value: 7 and @instance_ar value: 7
local_ar_2 value: 8 and local_ar value: 8 and @instance_ar value: 8

问题A:pushlocal_ar_2如何更改另外两个数组?我对局部变量的理解是,一旦它们被创建,它们不应该影响任何其他变量,因为它们是本地变量。

问题B:我怎样才能避免这样的事情发生? 来自C和Perl这只是让我大吃一惊。

2 个答案:

答案 0 :(得分:9)

Ruby使用引用!记在脑子里。如果你想要一份副本,你必须这样做:

@instance_ar = [1,2,3,4]
local_ar = @instance_ar.clone
local_ar_2 = local_ar.clone

编辑:

示例:

a = ["a", "b", "c"]
b = a[0]
b = "d" # We assign a new object to b!

a是:     => [“a”,“b”,“c”]

但:

a = ["a", "b", "c"]
b = a[0]
b[0] = "d" # We are working with the reference!

a is:
=> ["d", "b", "c"]

a = "hello"
b = a
b += " world" 
# Is the same as b = b + " world", we assign a new object!

a是:     => “你好”

但:

a = "hello"
b = a
b<<" world"
# We are working with the reference!

a是:     =&GT; “你好世界”

a = "abc"
b = a
b[0] = "d" # we are working with the reference

a是:     =&GT; “DBC”

您可以在此处阅读有关它的所有内容:http://ruby-doc.org/docs/ProgrammingRuby/。几乎在页面底部向下滚动到“变量”。

答案 1 :(得分:3)

在Ruby中,您不是“创建变量”,而是创建对象(在您的情况下为数组),并将它们分配给变量。因此,在您的示例中,您有一个包含两个名称的单个数组local_arlocal_ar_2。之后你改变了对象,而不是变量(它们都指向同一个改变的对象)。

你说你是来自C,看看这个例子:

int a[5];
int * b;

a[0] = 10;
a[1] = 20;
b = a;
b[1] = 5;

printf("%d", a[1]);

将打印什么号码?它与Ruby代码中发生的情况大致相同。

一个注意事项:使用for迭代数组在Ruby中不是惯用的。通常使用:

local_ar_2.each_with_index do |i, count|
    puts "local_ar_2 value: #{i} and local_ar value: #{local_ar[count]} and @instance_ar value: #{@instance_ar[count]}\n"
end