Explain me this line

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VB

pr.StartInfo.Arguments = i.SubItems(3).Text.Remove(i.SubItems(3).Text.LastIndexOf("\") + 1, _
                                                (i.SubItems(3).Text.Length - i.SubItems(3).Text.LastIndexOf("\") - 1))

The subitems(3).text is for exemple : C:\Program Files (x86)\Internet Download Manager\IDMAN.exe

Posted 22-Jun-13 11:10am

Haykerman

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CPallini · Accepted Answer · 2013-06-22T11:41:00

Solution 1

Removes the final part of the string, from the last occurring "\" character (excluded).
You may re-write it as

VB

s.Remove(s.LastIndexOf("\") + 1, s.Lenght - s.LastIndexOf("\") - 1)

where s.LastIndexOf("\") is the index of the last occurring "\" character, hence
s.Lenght -s.LastIndexOf("\")-1 is the number of characters following the last "\".

See String.Remove Method (Int32, Int32)[^].

Posted 22-Jun-13 11:41am

CPallini

Comments

Haykerman 23-Jun-13 9:05am

Thanks can you explain me : what is +1 and -1 ?

CPallini 23-Jun-13 12:54pm

You need +1 because you want to remove the part of the string succeeding the last "\" occurrence. Then you may find the same +1, negated, in the formula that computes the length of such part of the string.

Haykerman 23-Jun-13 14:57pm

Thanks!

CPallini 23-Jun-13 15:09pm

You are welcome.