Setting Class Methods Using Strings

[Demonic Software]

[Note: parts of this message were removed to make it a legal post.]

Hello,

First question:
With the 'print 'foo_method called with \#{arg}', is there a way to escape
the "#" so that arg will not be evaluated until after the statement is
evalled? For example:

Let c be "'print 'foo_method called with \#{arg}'"

eval(c) => print 'foo_method called with #{arg}'



Second (Main Question)
Is it possible to set an object's methods with raw code and a method name?
I am not sure how to Google for an example on this, so I will just show an
example.

== Here is an empty class definition Foo, and later on I will want to
assign/add/set a method in an instantiated Foo object ==

class Foo
end

== Now I want to set a method in an instantiated Foo with code, so that I
can call f ==

code = "def foo_method(arg)
print 'foo_method called with #{arg}'
end"

f = Foo.new()
#...
#code to set a method in f
#...

f.foo_method("for the win!")

==

Is this possible? or Do I need create a string and add the method in like
the following example?

code = "\ndef foo_method(arg)
print 'foo_method called with \#{arg}'
end\n"
foo = "class Foo " + code + " end"

eval(foo)
f = Foo.new()
f.foo_method("for the win!")



Thanks in advance for your help.

 

[Chris Shea]

[Note: parts of this message were removed to make it a legal post.]

Hello,

First question:
With the 'print 'foo_method called with \#{arg}', is there a way to escape
the "#" so that arg will not be evaluated until after the statement is
evalled? For example:

Let c be "'print 'foo_method called with \#{arg}'"

eval(c) => print 'foo_method called with #{arg}'

Second (Main Question)
Is it possible to set an object's methods with raw code and a method name?
I am not sure how to Google for an example on this, so I will just show an
example.

== Here is an empty class definition Foo, and later on I will want to
assign/add/set a method in an instantiated Foo object ==

class Foo
end

== Now I want to set a method in an instantiated Foo with code, so that I
can call f ==

code = "def foo_method(arg)
print 'foo_method called with #{arg}'
end"

f = Foo.new()
#...
#code to set a method in f
#...

f.foo_method("for the win!")

==

Is this possible? or Do I need create a string and add the method in like
the following example?

code = "\ndef foo_method(arg)
print 'foo_method called with \#{arg}'
end\n"
foo = "class Foo " + code + " end"

eval(foo)
f = Foo.new()
f.foo_method("for the win!")

Thanks in advance for your help.

If you know when to use single quotes and when to use double quotes,
yes to both:


a = 1
c = 'puts "a = #{a}"' # => "puts \"a = \#{a}\""
eval(c) # >> a = 1

code = 'def foo_method(arg); puts "foo_method called with #{arg}";
end'
class Foo; end
Foo.class_eval(code)
Foo.new.foo_method('hi') # >> foo_method called with hi


Single-quoted strings do not perform interpolation, while double-
quoted strings do. So if you juggle them for eval, you can have it do
what you want. Of course, putting a backslash before the #{} in double-
quoted strings will escape that so it doesn't perform the
interpolation.

Play around and you should get the hang of it quickly. And beware
eval. Use carefully.

HTH,
Chris

 

[Demonic Software]

[Note: parts of this message were removed to make it a legal post.]

Cool thanks Chris.

[Note: parts of this message were removed to make it a legal post.]

Hello,

First question:
With the 'print 'foo_method called with \#{arg}', is there a way to escape
the "#" so that arg will not be evaluated until after the statement is
evalled? For example:

Let c be "'print 'foo_method called with \#{arg}'"

eval(c) => print 'foo_method called with #{arg}'

Second (Main Question)
Is it possible to set an object's methods with raw code and a method name?
I am not sure how to Google for an example on this, so I will just show an
example.

== Here is an empty class definition Foo, and later on I will want to
assign/add/set a method in an instantiated Foo object ==

class Foo
end

== Now I want to set a method in an instantiated Foo with code, so that I
can call f ==

code = "def foo_method(arg)
print 'foo_method called with #{arg}'
end"

f = Foo.new()
#...
#code to set a method in f
#...

f.foo_method("for the win!")

==

Is this possible? or Do I need create a string and add the method in like
the following example?

code = "\ndef foo_method(arg)
print 'foo_method called with \#{arg}'
end\n"
foo = "class Foo " + code + " end"

eval(foo)
f = Foo.new()
f.foo_method("for the win!")

Thanks in advance for your help.

If you know when to use single quotes and when to use double quotes,
yes to both:


a = 1
c = 'puts "a = #{a}"' # => "puts \"a = \#{a}\""
eval(c) # >> a = 1

code = 'def foo_method(arg); puts "foo_method called with #{arg}";
end'
class Foo; end
Foo.class_eval(code)
Foo.new.foo_method('hi') # >> foo_method called with hi


Single-quoted strings do not perform interpolation, while double-
quoted strings do. So if you juggle them for eval, you can have it do
what you want. Of course, putting a backslash before the #{} in double-
quoted strings will escape that so it doesn't perform the
interpolation.

Play around and you should get the hang of it quickly. And beware
eval. Use carefully.

HTH,
Chris

 

[Todd Benson]

Hello,

First question:
With the 'print 'foo_method called with \#{arg}', is there a way to escape
the "#" so that arg will not be evaluated until after the statement is
evalled? For example:

Let c be "'print 'foo_method called with \#{arg}'"

eval(c) => print 'foo_method called with #{arg}'



Second (Main Question)
Is it possible to set an object's methods with raw code and a method name?
I am not sure how to Google for an example on this, so I will just show an
example.

== Here is an empty class definition Foo, and later on I will want to
assign/add/set a method in an instantiated Foo object ==

class Foo
end

== Now I want to set a method in an instantiated Foo with code, so that I
can call f ==

code = "def foo_method(arg)
print 'foo_method called with #{arg}'
end"

f = Foo.new()
#...
#code to set a method in f
#...

f.foo_method("for the win!")

==

Is this possible? or Do I need create a string and add the method in like
the following example?

code = "\ndef foo_method(arg)
print 'foo_method called with \#{arg}'
end\n"
foo = "class Foo " + code + " end"

eval(foo)
f = Foo.new()
f.foo_method("for the win!")



Thanks in advance for your help.

Unless you absolutely _have_ to create a string (like you're holding
code in a database that is to be executed)...

class Foo
end

f = Foo.new
g = Foo.new

def f.bar arg; puts arg; end

f.bar "hi"
# "hi"
g.bar "hi"
# undefined method error

It can also be done like this after the class construction...

class << f
def bar arg; puts arg; end
end

I don't think this case is a good use of #eval, but it depends on what
you're doing.

hth,
Todd