groovy

String containing an integer is considered a character for ranges

Details

  • Type: Bug Bug
  • Status: Closed Closed
  • Priority: Major Major
  • Resolution: Won't Fix
  • Affects Version/s: 1.0-JSR-4
  • Fix Version/s: 1.0-JSR-5
  • Component/s: None
  • Labels:
    None
  • Number of attachments :
    0

Description

Let's say I have a method that receives a string that contains an integer (0..9). If I use that parameter as part of a range, Groovy considers it a character which causes some really unexpected effects. For example:

str = "5"
// ...
lastNumber = str;
myRange = 0..lastNumber
myRange.each {
print it
}

I would expect it to print: 012345
Instead it takes the character value of 5 and prints: 0123456789101112 ... 515253

Is this what is really wanted or should this be considered a bug? At least it makes it pretty easy to make mistakes.

Activity

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Markus Halttunen added a comment -

Actually this is not specific to ranges. Groovy seems to consider a string whose length is one as a character. I guess there is nothing that can be done about that? I would just have to say: "lastNumber = str.toInteger()"?

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Markus Halttunen added a comment - Actually this is not specific to ranges. Groovy seems to consider a string whose length is one as a character. I guess there is nothing that can be done about that? I would just have to say: "lastNumber = str.toInteger()"?
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blackdrag blackdrag added a comment -

Strings of length 1 are Strings in the first line, they are converted to chars if needed and as such htey are also a number with the ordinal value of the char. You mix 2 types, String and Integer. Using "0" as start for the Range is on possibility if it is ok to use a StringRange, or you can use str.toInteger() as you mentioned by yourself

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blackdrag blackdrag added a comment - Strings of length 1 are Strings in the first line, they are converted to chars if needed and as such htey are also a number with the ordinal value of the char. You mix 2 types, String and Integer. Using "0" as start for the Range is on possibility if it is ok to use a StringRange, or you can use str.toInteger() as you mentioned by yourself

People

  • Assignee:
    blackdrag blackdrag
    Reporter:
    Markus Halttunen
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Dates

  • Created:
    Updated:
    Resolved: