p "4".to_i +3    # => 0 这个结果很费解,竟然不是 7
p "4".to_i + 3   # => 7
p 4 +3             # => 7
p 4 + 3            # => 7

如果不开终端确认，第一个结果我就搞错了。我知道这是错误的省略空格导致的歧义，可是为什么会得零？ 尝试获取语法树的结果：

require 'parser/current'
# opt-in to most recent AST format:
Parser::Builders::Default.emit_lambda   = true
Parser::Builders::Default.emit_procarg0 = true
Parser::Builders::Default.emit_encoding = true
Parser::Builders::Default.emit_index    = true
pp Parser::CurrentRuby.parse('"4".to_i +3')
pp Parser::CurrentRuby.parse('"4".to_i + 3')
pp Parser::CurrentRuby.parse('"4".to_i +3').loc
pp Parser::CurrentRuby.parse('"4".to_i + 3').loc

输出为

s(:send,
  s(:str, "4"), :to_i,
  s(:int, 3))
s(:send,
  s(:send,
    s(:str, "4"), :to_i), :+,
  s(:int, 3))
#<Parser::Source::Map::Send:0x00007f9ff589c348
 @begin=nil,
 @dot=#<Parser::Source::Range (string) 3...4>,
 @end=nil,
 @expression=#<Parser::Source::Range (string) 0...11>,
 @node=s(:send,
  s(:str, "4"), :to_i,
  s(:int, 3)),
 @selector=#<Parser::Source::Range (string) 4...8>>
#<Parser::Source::Map::Send:0x00007f9ff78428c8
 @begin=nil,
 @dot=nil,
 @end=nil,
 @expression=#<Parser::Source::Range (string) 0...12>,
 @node=s(:send,
  s(:send,
    s(:str, "4"), :to_i), :+,
  s(:int, 3)),
 @selector=#<Parser::Source::Range (string) 9...10>>

没看出操作符是怎么丢失的，再底层一些：

require 'ripper'
pp Ripper.sexp '"4".to_i + 3'
pp Ripper.sexp '"4".to_i +3'

输出为：

[:program,
 [[:binary,
   [:call,
    [:string_literal, [:string_content, [:@tstring_content, "4", [1, 1]]]],
    :".",
    [:@ident, "to_i", [1, 4]]],
   :+,
   [:@int, "3", [1, 11]]]]]
[:program,
 [[:command_call,
   [:string_literal, [:string_content, [:@tstring_content, "4", [1, 1]]]],
   :".",
   [:@ident, "to_i", [1, 4]],
   [:args_add_block, [[:@int, "+3", [1, 9]]], false]]]]

仍然没看出为何会得零？