Hello Bret,
Friday, November 15, 2002, 3:01:26 AM, you wrote:
foo ||= “oink”
foo = (foo || “oink”)
from the “programming ruby”:
Variable/Method Ambiguity
When Ruby sees a name such as ``a’’ in an expression, it needs to determine if it is a local variable reference or a call to a method with no parameters. To decide which is the case, Ruby uses a heuristic. As Ruby reads a source file, it keeps track of symbols that have been assigned to. It assumes that these symbols are variables. When it subsequently comes across a symbol that might be either a variable or a method call, it checks to see if it has seen a prior assignment to that symbol. If so, it treats the symbol as a variable; otherwise it treats it as a method call. As a somewhat pathological case of this, consider the following code fragment, submitted by Clemens Hintze.
def a
print “Function ‘a’ called\n”
99
end
for i in 1…2
if i == 2
print “a=”, a, “\n”
else
a = 1
print “a=”, a, “\n”
end
end
produces: a=1
Function ‘a’ called
a=99
During the parse, Ruby sees the use of a'' in the first print statement and, as it hasn't yet seen any assignment to a,‘’ assumes that it is a method call. By the time it gets to the second print statement, though, it has seen an assignment, and so treats ``a’’ as a variable.
Note that the assignment does not have to be executed—Ruby just has to have seen it. This program does not raise an error.
a = 1 if false; a
···
–
Best regards,
Bulat mailto:bulatz@integ.ru