I understand the following:

$ ruby -e "p 'abc'.sub('a','A').sub('b', 'B')"
"ABc"

I am OK with the following also:

echo abc | ruby -p -e "sub('a','A');sub('b', 'B')"
ABc

But:

echo abc | ruby -p -e "sub('a','A').sub('b', 'B')"
Abc

I expect the result to be "ABc" as well, why is it not? The second sub('b', 'B') is not operational.

CodePudding user response：

The two cases look similar, but in fact you are running different methods from the Ruby core library in them:

In your first case, i.e. sub('a','A');sub('b', 'B'):

You are running both sub without specifying an explicit receiver, and therefore you are invoking the method Kernel#sub. The Ruby-Doc says about this method:

sub(pattern, replacement) → $_

Equivalent to $_.sub(args), except that $_ will be updated if substitution occurs. Available only when -p/-n command line option specified.

Hence, in the first example, you really invoke that Kernel#sub twice, and after each invocation, $_ is updated. Therefore, $_ is ABc after the second sub has been executed. At the end of the of the whole expression supplied by -e (i.e. at the end of the implicit loop provided by the -p option), the value of $_ is printed, and you see ABc.

In your second example, i.e.

sub('a','A').sub('b', 'B')

The first sub again is Kernel#sub, as before. It has the effect of turning the string into Abc, and also sets $_ to Abc. However, the second sub now does have an explicit receiver (the string resulting from the first sub), and in this case, the method String#sub is executed. This method produces ABc, but different to Kernel#sub, it does not update $_. Therefore, $_ is still set to Abc, and this is what you see as output.