Question

from Programming Perl pg 90, he says:

   @ary = (1, 3, sort 4, 2);
    print @ary;

the commas on the right of the sort are evaluated before the sort but the commas on the left are evaluated after. … list operators tend to gobble .. and then act like a simple term”

1. Does the assignment result in sort being processed or does that happen when @ary is expanded by print?
2. What does he mean by all that “comma” stuff?? My understanding is that in the assignment statement, comma has a lower priority than a list operator therefore sort runs first and gobbles up it’s arguments (4 and 2).. How the heck is comma being evaluated at all?? So that statemnent then becomes (1, 3, 2, 4) a list which is assigned.. comma is just acting as a list separator and not an operator!! In fact on pg:108 he says: do not confuse the scalar context use of comma with list context use..

3. What is a leftward and rightward list operator? print @ary is a rightward list operator?? So it has very low priority?

   print($foo, exit);

here, how is precedence evaluated? print is a list operator that looks like a function so it should run first! it has two arguments $foo and exit.. so why is exit not treated as a string??? After all priority-wise print(the list operator) has higher priority??

print $foo, exit;

here, you have print and , operators but the list operator has higher precedence.. so.. exit should be treated as a string – why not??

   print ($foo & 255) + 1, "\n";

here since it’s a list operator it prints $foo & 255 Shouldn’t something similar happen with the above mentioned exit stuff..

Answer 1

When in doubt about how Perl is parsing a construct, you can run the code through the B::Deparse module, which will generate Perl source code from the compiled internal representation. For your first example:

$ perl -MO=Deparse,-p -e '@ary = (1, 3, sort 4, 2); print @ary;'
(@ary = (1, 3, sort(4, 2)));
print(@ary);
-e syntax OK

So as you can see, sort takes the two arguments to its right.

As far as execution order goes, you can find that out with the B::Concise module (I’ve added the comments):

$ perl -MO=Concise,-exec -e '@ary = (1, 3, sort 4, 2); print @ary;'
1  <0> enter 
2  <;> nextstate(main 1 -e:1) v:{
3  <0> pushmark s       # start of list
4  <$> const[IV 1] s    # 1 is added to list
5  <$> const[IV 3] s    # 3 is added to list
6  <0> pushmark s       # start of sort's argument list
7  <$> const[IV 4] s    # 4 is added to sort's argument list
8  <$> const[IV 2] s    # 2 is added to sort's argument list
9  <@> sort lK          # sort is run, and returns its list into the outer list
a  <0> pushmark s
b  <#> gv[*ary] s
c  <1> rv2av[t2] lKRM*/1
d  <2> aassign[t3] vKS/COMMON  # the list is assigned to the array
e  <;> nextstate(main 1 -e:1) v:{
f  <0> pushmark s         # start of print's argument list
g  <#> gv[*ary] s         # the array is loaded into print's argument list
h  <1> rv2av[t5] lK/1
i  <@> print vK           # print outputs it's argument list
j  <@> leave[1 ref] vKP/REFC
-e syntax OK

For your second example:

$ perl -MO=Deparse,-p -e 'print $foo, exit;'
print($foo, exit);
-e syntax OK

$ perl -MO=Concise,-exec -e 'print $foo, exit;'
1  <0> enter 
2  <;> nextstate(main 1 -e:1) v:{
3  <0> pushmark s
4  <#> gvsv[*foo] s   # add $foo to the argument list
5  <0> exit s         # call `exit` and add its return value to the list
6  <@> print vK       # print the list, but we never get here
7  <@> leave[1 ref] vKP/REFC
-e syntax OK

So as you can see, the exit builtin is run while trying to assemble the argument list for print. Since exit causes the program to quit, the print command never gets to run.

And the last one:

$ perl -MO=Deparse,-p -e 'print ($foo & 255) + 1, "\n";'
((print(($foo & 255)) + 1), '???');  # '???' means this was optimized away
-e syntax OK

$ perl -MO=Concise,-exec -e 'print ($foo & 255) + 1, "\n";'
1  <0> enter 
2  <;> nextstate(main 1 -e:1) v:{
3  <0> pushmark v
4  <0> pushmark s
5  <#> gvsv[*foo] s
6  <$> const[IV 255] s
7  <2> bit_and[t2] sK
8  <@> print sK
9  <$> const[IV 1] s
a  <2> add[t3] vK/2
b  <@> list vK
c  <@> leave[1 ref] vKP/REFC
-e syntax OK