块定义 - 大括号和 do-end 之间的区别? [英] Block definition - difference between braces and do-end?
问题描述
任何人都可以解释为什么以下代码会因错误而中止
can anybody explain why the following code aborts with an error
irb(main):186:0> print ((1..10).collect do |x| x**2 end)
SyntaxError: (irb):186: syntax error, unexpected keyword_do_block,
expecting ')'
print ((1..10).collect do |x| x**2 end)
^
(irb):186: syntax error, unexpected keyword_end, expecting $end
print ((1..10).collect do |x| x**2 end)
^
from /usr/bin/irb:12:in `<main>'
而以下具有相同功能的代码是否按预期工作?
whereas following code with the same functionality works as expected ?
irb(main):187:0> print ((1..10).collect { |x| x**2 })
[1, 4, 9, 16, 25, 36, 49, 64, 81, 100]=> nil
我确实相信花括号{}"可以在块中任意替换do end"定义.
I did believed curly-braces "{ }" can substitute "do end" arbitrarily at block definition.
我知道我可以通过省略打印之间的空格来修复"第一个调用方法和第一个括号,但我对解释感兴趣为什么它的行为不同.
I know I can "fix" the first call by omitting a space between print method and the first parenthesis, but I'm interested in an explanation why it behaves different.
推荐答案
区别在于优先级:
# Equivalent to puts( (1..10).map { |i| i*2 } )
> puts (1..10).map { |i| i*2 }
2
4
6
8
10
12
14
16
18
20
=> nil
# Equivalent to puts( (1..10).map ) { |i| i*2 }
> puts (1..10).map do |i| i*2 end
#<Enumerator:0x928f24>
=> nil
在第一种情况下,块被传递给map
,一切正常.在第二种情况下,块被传递给 puts
,它不会对它做任何事情.map
不接收块,只返回一个枚举器.
In the first case, the block is passed to map
, and everything works properly.
In the second case, the block is passed to puts
, which doesn't do anything with it. map
doesn't receive a block and just returns an enumerator.
至于语法错误,如果你去掉print
和(
之间的空格,一切正常;)
As for the syntax error, if you remove the space between print
and (
everything works ;)
区别在于 ruby 是将括号视为方法参数分隔符,还是将其视为通用语句分组.我不确定那里的确切区别,但它微妙而烦人
The difference is whether ruby is treating your parentheses as method argument delimiters, or whether it's a generic statement grouping. I'm not sure of the exact difference there but it's subtle and annoying
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