Re: '...' is a variable,who can explain
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- Subject: Re: '...' is a variable,who can explain
- From: "Aaron Brown" <arundelo@...>
- Date: 2007年1月12日 11:55:14 -0500
"..." has two related meanings. If used in a function's
list of formal parameters, it allows the function to take an
indefinite number of arguments (up to 200):
return function(...)
Such a function is called a vararg function.
"..." can also be used as an expression (but not a variable,
which is another kind of expression). "..." used in this
way is called a vararg expression:
return f1(f2(...)) --...treated as a variable! is it maze!
Vararg expressions were introduced in Lua 5.1. They can
only be used if the enclosing function is a vararg function.
Like function calls, but unlike other expressions, vararg
expressions can evaluate to more -- or less -- than one
value, and are adjusted to one value in certain contexts.
function foo(...)
print(type((...)))
print(type(...))
end
foo(42, "ignored")
number
number
foo()
nil
stdin:3: bad argument #1 to 'type' (value expected)
stack traceback:
[C]: in function 'type'
stdin:3: in function 'foo'
stdin:1: in main chunk
[C]: ?
For more detail on multiple-valued expressions and their
adjustment, and on vararg functions and vararg expressions,
see:
http://www.lua.org/manual/5.1/manual.html#2.5
http://www.lua.org/manual/5.1/manual.html#2.5.9
--
Aaron
Beginning Lua Programming: http://www.amazon.com/gp/product/0470069171/