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The Ruby Language FAQ: Syntax Next Previous Contents

6. Syntax

6.1 What does :var mean?

A colon followed by a name generates an integer(Fixnum) called a symbol which corresponds one to one with the identifier. "var".intern gives the same integer as :var, but the ``:'' form will create a local symbol if it doesn't already exist.

The routines "catch", "throw", "autoload", and so on, require a string or a symbol as an argument.

"method_missing", "method_added" and "singleton_method_added" (and others) require a symbol.

The fact that a symbol springs into existence the first time it is referenced is sometimes used to assign unique values to constants:


6.2 How can I access the value of a symbol?

To get the value of the variable corresponding to a symbol, you can use id2name to get the name of the variable, and then eval that to get that variable's contents. In the scope of "symbol", do eval(:symbol.id2name).

a = 'This is the content of "a"'
b = eval(:a.id2name) ==  # b now references the same object as a

If your symbol corresponds to the name of a method, you can use the Method.method function to return a corresponding Method object, which you may then call.

class Demo
  def meth
    "Hello, world"

d =         # -> #<Demo:0x401b498c>
m = d.method(:meth)  # -> #<Method: Demo(Demo)#meth>               # -> "Hello, world"

6.3 Is loop a control structure?

Although loop looks like a control structure, it is actually a method defined in Kernel. The block which follows introduces a new scope for local variables.

6.4 Ruby doesn't have a post-test loop

Q: Ruby does not have a do { ... } while construct, so how can I implement loops that test the condition at the end.

Clemens Hintze says: You can use a combination of Ruby's begin ... end and the while or until statement modifiers to achieve the same effect:

i = 0
  puts "i = #{i}"
  i += 1
end until i > 4


i = 0
i = 1
i = 2
i = 3
i = 4

6.5 a +b gives an error!

Ruby works hard to distinguish method calls from operators, and variable names from method names. Unfortunately, there's no way it can get it right all the time. In this case, ``a +b'' is parsed as ``a(+b)''. Remove the space to the left of ``+'' or add a space to the right of ``+,'' and it will be parsed as an addition.

6.6 s = "x"; puts s *10 gives an error.

Again, Ruby sees the asymmetrical space and parses it as puts(s(*10)) (which isn't too smart, really). Use ``s*10'' or ``s * 10'' to get the desired result.

6.7 Why can't I pass a hash literal to a method: p {}?

The {} is parsed as a block, not a Hash constructor. You can force the {} to be treated as an expression by making the fact that it's a parameter explicit: p({}).

6.8 I can't get def pos=(val) to work.

I have the following code, but I cannot use the method pos = 1.

def pos=(val)
   print @pos, "\n"
   @pos = val

Methods with = appended must be called with a receiver (without the receiver, you're just assigning to a local variable). Invoke it as self.pos = 1.

6.9 What is the difference between '\1' and '\\1'?

They have the same meaning. In a single quote string, only \' and \\ are transformed and other combinations remain unchanged.

However, in a doubled quoted string, "\1" is the byte \001, while "\\1" is the two character string containing a backslash and the character "1".

6.10 What's the difference between ``or'' and ``||''?

Q: ``p(nil || "Hello")'' prints "Hello", while ``p(nil or "Hello")'' gives a parse error.

A: || combines terms within an expression. Because the first term in this case is nil, the second term is evaluated.

or is used to combine expressions in conditionals. Ruby is not expecting a conditional statement in an argument list.

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