Perl: while with no conditional

Question

According to the doc, the while statement executes the block as long as the expression is true. I wonder why it becomes an infinite loop with an empty expression:

Answer 1

This is a special case. An empty condition expression defaults to just true, which means "loop forever, or until a break. In C (and perl) the idiom

for(;;) {
   // Neverending fun
}

has the same effect for the same reason.

There doesn't appear to be any mention of this in the official perl docs, and yet there is a special rule in the parser for it. Perhaps it's because nobody uses it :)

The for(;;) idiom is less uncommon though.

Answer 2

This is a special case of the concept of Vacuous Truth. If there is no condition, the statement while the condition is true is itself vacuously true.

If I am reading this correctly, the relevant piece of code seems to be around line 5853 of op.c in 5.14.1:

5853     if (expr) {
5854         scalar(listop);
5855         o = new_logop(OP_AND, 0, &expr, &listop);
5856         if (o == expr && o->op_type == OP_CONST && !SvTRUE(cSVOPo->op_sv)) {
5857             op_free(expr);              /* oops, it's a while (0) */
5858             op_free((OP*)loop);
5859             return NULL;                /* listop already freed by new_logop */
5860         }
5861         if (listop)
5862             ((LISTOP*)listop)->op_last->op_next =
5863                 (o == listop ? redo : LINKLIST(o));
5864     }
5865     else
5866         o = listop;

I am assuming with no expr in the condition, we reach o = listop. listop was previously defined as listop = op_append_list(OP_LINESEQ, block, cont);.

Answer 3

$ perl -MO=Deparse -e 'while () { }'
while (1) {
    ();
}
-e syntax OK

It seems that while () {} and while (1) {} are equivalent. Also note that empty parens* are inserted in the empty block.

Another example of pre-defined compiler behaviour:

$ perl -MO=Deparse -e 'while (<>) { }'
while (defined($_ = <ARGV>)) {
    ();
}
-e syntax OK

I would say that this is just the docs not reporting a special case.

* — To be precise, the stub opcode is inserted. It does nothing, but serves a goto target for the enterloop opcode. There's no real reason to note this. Deparse denotes this stub op using empty parens, since parens don't generate code.