How to use compare on a version number where theres less parts in one number in Objective-C?

Question

I\'ve found the following code at http://snipplr.com/view/2771

Which is pretty good, almost exactly what I was looking for, but if I use the values @\"1.4.5\", @

Answer 1

My sense would be that the first numeral group is always the most significant, so 10.anything is bigger than 9.anything.anything. If I'm right about that, then the solution is to replace the dots with zeros and pad the shorter string on the right-hand side with zeros to match the length of the longer string:

e.g.
9.4   --->  90400  (padded on the right with 00)
8.6.7 --->  80607

What's nice about this is, that if I'm wrong about the requirement, the algorithm can be readily fixed by padding the shorter string on the right.

- (NSComparisonResult)compareVersion:(NSString *)vA withVersion:(NSString *)vB {

    NSString *vAPadded = [vA stringByReplacingOccurrencesOfString:@"." withString:@"0"];
    NSString *vBPadded = [vB stringByReplacingOccurrencesOfString:@"." withString:@"0"];

    while (vAPadded.length < vBPadded.length)
        vAPadded = [vAPadded stringByAppendingString:@"0"];

    while (vBPadded.length < vAPadded.length)
        vBPadded = [vBPadded stringByAppendingString:@"0"];

    return [vAPadded intValue] - [vBPadded intValue];
}

If I've got the requirement backwards on significant digits, change the pads like this:

vAPadded = [@"0" stringByAppendingString:vAPadded];