[SOLVED] How does whitespace IFS function?

glenjoker · 10-13-2015, 08:11 AM

I just read from somewhere that if I had IFS set to be colon (i.e. 'IFS=:'), then the string "a:b:: c d "(2 spaces in between 'c' and 'd') would be separated into 4 fields, namely "a","b",""," c d ", which is quite obvious. However, it then said that if IFS was changed to be colon and space (i.e. 'IFS= :'), then the leading and trailing space of the string would be ignore and consecutive spaces would function as a separator as well; therefore, the same string, "a:b:: c d ", would be separated into 5 fields, "a","b","","c" and "d", but I think that it should be 6 fields, "a","b",""(this null character field is the one in between the two colons), ""(this null character field is the one in between the colon and the space, which is now a separator),"c","d". Which one is correct?

grail · 10-13-2015, 11:40 AM

Quote:

but I think that it should be

Sad to say, but irrelevant of what you may think, if you set IFS as described with that input you cannot argue with the output. So there is no question on which is correct.

So I think the part of man that explains what you see is in red below:

Code:

Word Splitting
       The shell scans the results of parameter expansion, command substitution, and arithmetic expansion that did not occur within double quotes for word splitting.

       The  shell  treats  each  character  of  IFS  as a delimiter, and splits the results of the other expansions into words using these characters as field terminators.  If IFS is unset, or its value is
       exactly <space><tab><newline>, the default, then sequences of <space>, <tab>, and <newline> at the beginning and end of the results of the previous expansions are ignored, and any  sequence  of  IFS
       characters not at the beginning or end serves to delimit words.  If IFS has a value other than the default, then sequences of the whitespace characters space and tab are ignored at the beginning and
       end of the word, as long as the whitespace character is in the value of IFS (an IFS whitespace character).  Any character in IFS that is not IFS whitespace, along with any  adjacent  IFS  whitespace
       characters, delimits a field.  A sequence of IFS whitespace characters is also treated as a delimiter.  If the value of IFS is null, no word splitting occurs.

glenjoker · 10-14-2015, 01:21 AM

Quote:

Originally Posted by grail

Sad to say, but irrelevant of what you may think, if you set IFS as described with that input you cannot argue with the output. So there is no question on which is correct.
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This reminds me of my Electrodynamics professor who once said the same to me, but as I am very so often nowhere near uncovering or discerning the truth, what else can I do but have my superficial thoughts expressed and consult the experts so that I can learn from my mistakes.

Anyway, thanks for replying. It really clears up my doubts, and I really appreciate it.