Re^7: Why is const x const not a const?

Sorry, there was a set of parens missing. (Fixed.) I was about to reply to your post that for (CONST..CONST) is a counting loop rather than a loop iterating over the result of a range operator. There is no range operator, so the range operator cannot be flattened. Try

perl -e"<>; exit; @a = 1..100000;"
perl -e"<>; exit; for ((), 1..100000) {}"
[download]

Results:

perl -e"<>; exit; for ((),1..   1000) {}    1,252K
perl -e"<>; exit; for ((),1..  10000) {}    1,784K
perl -e"<>; exit; for ((),1.. 100000) {}    6,028K
perl -e"<>; exit; for ((),1..1000000) {}   45,212K
[download]

Update: Added results.

Comment on Re^7: Why is const x const not a const? Select or Download Code

Replies are listed 'Best First'.
Re^8: Why is const x const not a const? by BrowserUk (Patriarch) on Jan 23, 2012 at 19:31 UTC
You are forcing the issue there with the otherwise pointless parens. With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday' Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error. "Science is about questioning the status quo. Questioning authority". In the absence of evidence, opinion is indistinguishable from prejudice. The start of some sanity?	[reply]
Re^9: Why is const x const not a const? by ikegami (Patriarch) on Jan 23, 2012 at 22:12 UTC
Yes, I forced the use of the range operator. After all, we are talking about the range operator. I didn't say «`for (CONST..CONST)`» flattens; I said «`CONST..CONST`» does, and I was referring to the range operator.	[reply] [d/l] [select]