using ruby string interpolation while mutating object

Question

Just encountered some strange behaviour while using string interpolation to print an object and his value after being passed to a function.

print "#{obj} => #{function obj}"

It outputed something similar both for #{obj} and #{function obj}

I came to the conclusion that it was because my function was a mutating function. I can reproduce the output easily with :

def mutate_the_object(obj)
  obj.concat '1'
end
obj = '1'
print "#{obj} != #{mutate_the_object obj} != #{mutate_the_object obj} != #{obj}"

=> 111 != 111 != 111 != 111

Why does the string interpolation give 4 equal fields? I thought the result would be something like :

=> 1 != 11 != 111 != 111

Obviously, because the string interpolation has the lowest precedence ever. — Aleksei Matiushkin
– Aleksei Matiushkin, Commented Apr 13, 2018 at 12:09
@mudasobwa yes I did no think of that, in my mind it had to work like it would in C using printf("%d %d %d",i,++i,++i); thanks — Alexis Delahaye
– Alexis Delahaye, Commented Apr 13, 2018 at 12:20
@AlexisDelahaye it’s like "%s != %s != %s != %s" % [o, m(o), m(o), o]. First the RHO array is prepared (both calls to m(o) are done,) then the string is being interpolated. — Aleksei Matiushkin
– Aleksei Matiushkin, Commented Apr 13, 2018 at 12:42
All arguments in RHO are the very same object. def m(o); o << '1'; end; o = '1'; [o, m(o), m(o), o].map(&:__id__) #⇒ [46935826870120, 46935826870120, 46935826870120, 46935826870120]. That is the same object. — Aleksei Matiushkin
– Aleksei Matiushkin, Commented Apr 13, 2018 at 12:47
TL;DR: mutation is a base ground of 99% of bugs and/or unmaintainable code. — Aleksei Matiushkin
– Aleksei Matiushkin, Commented Apr 13, 2018 at 12:48

axiac · Accepted Answer · 2018-04-13 12:58:27Z

1

String#concat changes the object and returns it.

def mutate_the_object changes the object it receives as argument and returns it.

It seems the interpolated expressions are computed first, then their values are used to compute the string, in a similar fashion the C function sprintf() works.

The code posted in the question is equivalent to:

def mutate_the_object(obj)
  obj.concat '1'
end
obj = '1'
obj1 = mutate_the_object obj
obj2 = mutate_the_object obj
print "#{obj} != #{obj1} != #{obj2} != #{obj}"

obj1 and obj2 are variables that hold references to the same object as obj.

This can be easily proved by running:

print <<END
    #{obj.__id__}
    #{(mutate_the_object obj).__id__}
    #{(mutate_the_object obj).__id__}
    #{obj.__id__}
END

answered Apr 13, 2018 at 12:58

axiac

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Comments

Jörg W Mittag · Accepted Answer · 2018-04-13 12:41:31Z

I looked into the ISO Ruby Language Specification, the RubySpec, and the RDoc for interpolated String literals, and neither of the three define a specific evaluation order for interpolations. They probably should explicitly say that evaluation-order is undefined, but still, the fact that it isn't explicitly defined, means that it is implicitly undefined.

So, in other words: if your interpolations have side-effects, the order of those side-effects both with respect to each other and the time when they are inserted into the string is unspecified, and what you are observing is a perfectly legal implementation of the spec.

From what I can tell, (at least) any of the following results is legal:

'1 != 11 != 111 != 1'
'1 != 11 != 111 != 11'
'1 != 11 != 111 != 111'
'1 != 111 != 11 != 1'
'1 != 111 != 11 != 11'
'1 != 111 != 11 != 111'
'1 != 111 != 111 != 1'
'1 != 111 != 111 != 11'
'1 != 111 != 111 != 111'
'11 != 11 != 111 != 1'
'11 != 11 != 111 != 11'
'11 != 11 != 111 != 111'
'11 != 111 != 11 != 1'
'11 != 111 != 11 != 11'
'11 != 111 != 11 != 111'
'11 != 111 != 111 != 1'
'11 != 111 != 111 != 11'
'11 != 111 != 111 != 111'
'111 != 11 != 111 != 1'
'111 != 11 != 111 != 11'
'111 != 11 != 111 != 111'
'111 != 111 != 11 != 1'
'111 != 111 != 11 != 11'
'111 != 111 != 11 != 111'
'111 != 111 != 111 != 1'
'111 != 111 != 111 != 11'
'111 != 111 != 111 != 111'

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