Sebagai pegolf kode, kami tidak terbiasa melepaskan ( pasti ). Kami akan membutuhkan beberapa alat untuk membantu kami melakukan itu.

Tentu saja, untuk membantu memasarkan rilis baru, kami membutuhkan Versi Rilis yang bagus dan berkilau. Siapa yang tidak senang ketika mereka mendengar tentang versi 3.0.0?

Tugas

Tugas Anda adalah menulis program / rutin / ... untuk menambah nomor versi.

Anda perlu menambah nomor versi dan mengatur ulang yang "kurang penting" (yaitu versi tambalan).

Anda mendapatkan dua argumen: versi saat ini (ex "1.0.3") sebagai string, dan indeks untuk mengetahui mana yang akan diperbarui (0 atau 1-diindeks).

Contoh, diindeks 0:

next-version("1.0.3", 0) # 2.0.0
next-version("1.2.3.4.5", 2) # 1.2.4.0.0
next-version("10.0", 0) # 11.0
next-version("3", 0) # 4
next-version("1", 7) # ERROR
next-version("01", 0) # ERROR

Versi ini adalah string, setiap bagian adalah angka, dipisahkan dengan titik. Tidak ada titik awal, tidak ada jejak atau tidak ada titik berurutan (dan tidak ada di luar angka / titik). Tidak ada batasan ukuran string versi.

^[1-9]\d*(\.[1-9]\d*)*$

Kasus kesalahan (dua contoh terakhir) adalah perilaku yang tidak terdefinisi. Apa yang terjadi jika input yang salah tidak ada hubungannya dengan tantangan ini.

Seperti biasa, lubang standar dilarang. Anda diizinkan mencetak atau mengembalikan string.

Japt, 16 11 byte

¡V«´V+ÂX}'.

Uji secara online! Nomor input diindeks 1.

Berdasarkan jawaban JavaScript saya. Ini mengambil keuntungan dari salah satu fitur Japt yang paling membantu: memisahkan satu string pada yang lain sebelum memetakan setiap item, kemudian bergabung pada string itu lagi setelah memetakan.

Tanpa penjelasan dan penjelasan

¡  V«  ´ V+Â X}'.
Um@V&&!--V+~~X}'.
                   Implicit: U = input string, V = input number
Um@           }'.  Split U at periods, then map each item X by this function:
   V&&               If V is 0, return V.
      !--V+~~X       Otherwise, decrement V and return X + !V (increments X iff V was 1).
               '.  Re-join the result with periods.
                   Implicit: output last expression

Vim 20 25 byte

Sayangnya saya menyadari bahwa itu tidak menangani kasus memperbarui nomor terakhir, jadi saya harus menambahkan byte. Ini 1-diindeks.

DJA.0@"t.qq2wcw0@qq@qx

TryItOnline

Tidak dapat dicetak:

DJA.^[0@"t.^Aqq2wcw0^[@qq@qx

Ini mengambil argumen dalam urutan terbalik, sebagai baris terpisah:

3
1.2.3.4.5

Penjelasan:

DJ                           # Delete the input argument, and join lines
  A.^[0                      # Add a period to the end
       @"t.                  # Move to the "Copy register"th period
           ^A                # Increment the number under the cursor
             qq       @qq@q  # Until an error
               2w            # Move the cursor forward to the next number
                 cw0^[       # Change the number to 0
                           x # Delete the last '.'

JavaScript (ES6), 44 42 40 37 byte

Disimpan 3 byte berkat @Neil

x=>i=>x.replace(/\d+/g,n=>i&&+n+!--i)

Nomor input diindeks 1.

Cuplikan tes

f = x=>i=>x.replace(/\d+/g,n=>i&&+n+!--i)

console.log(f("1.0.3")(1))
console.log(f("1.2.3.4.5")(3))
console.log(f("10.0")(1))
console.log(f("1")(8))

V , 13 , 12 byte

Àñf.ñò2wcw0

Cobalah online!

Ini diindeks 0.

Ada ctrl-a(ASCII 0x01) di sana, jadi di sini adalah versi yang dapat dibaca:

Àñf.ñ<C-a>ò2wcw0

Penjelasan:

À                   " 'arg1' times
 ñ  ñ               " Repeat the following:
  f.                "   Move to the next '.' character
     <C-a>          " Increment the next number
          ò         " Recursively:
           2w       "   Move two words forward
              cw    "   Change this word
                0   "   to a '0'

Perl, 40 37 34 + 1 = 35 byte

-2 byte terima kasih kepada @Dada. -3 byte berkat ide yang saya dapatkan dari membaca kode Japt @ ETHproductions.

Jalankan dengan -pbendera.

$a=<>;s/\d+/$a--<0?0:$&+!($a+1)/ge

Cobalah online!

Rincian kode

-p          #Wraps the program in a while(<>){ ... print$_} statement.
            #Input is read into the $_ variable
$a=<>;s/\d+/$a--<0?0:$&+!($a+1)/ge
$a=<>;                              #Reads the version update into $a
      s/   /                   /ge  #Substitution regex:
                                g   #Repeats the substitution after first match
                                 e  #Interprets replacement as Perl code
       \d+                          #Matches 1 or more digits, and stores match in $&
                  ? :               #Ternary operation
            $a--<0                  #Checks if $a is negative and then decrements $a
                  ?0                #If negative, we've passed our match; print 0 instead
                    :$&+!($a+1)     #Otherwise, we're either printing $& + 0 (if $a was positive) or $& + 1 (if $a was 0).
#Since substitution implicitly modifies $_, and -p prints $_, it prints our answer

Jelly , 19 17 byte

ṣ”.V€‘⁹¦µJ’<⁹×j”.

1-diindeks.

TryItOnline!

Bagaimana?

ṣ”.V€‘⁹¦µJ’<⁹×j”. - Main link: currentVersion, theIndex
ṣ”.               - ṣplit left (currentVersion) at occurences of '.'
   V€             - eVal €ach (creates a list of integers)
      ⁹           - right argument (theIndex)
       ¦          - apply to given index(es)
     ‘            -    increment
        µ         - monadic chain separation (call the above result z)
         J        - range(length(z))  i.e. ([1,2,3,...,length])
          ’       - decrement         i.e. ([0,1,2,...,length-1])
            ⁹     - right argument (theIndex)
           <      - less than?        i.e. ([1,1,...,(1 at theIndex),0...,0,0,0]
             ×    - multiply by z (vectortises) - zeros out all of z after theIndex
              j”. - join with '.'

MATLAB, 85 byte

function f(s,j);a=split(s,'.');a(j)=string(double(a(j))+1);a(j+1:end)='0';join(a,'.')

Satu upaya berbasis golf pertama!

C# 116 104 Bytes

using System.Linq;(v,i)=>string.Join(".",v.Split('.').Select(int.Parse).Select((x,n)=>n==i?x+1:n>i?0:x));

Explanation

using System.Linq;(v,i)=>   //Anonymous function and mandatory using
    string.Join(".",                    //Recreate the version string with the new values
        v.Split('.')                    //Get individual pieces
            .Select(int.Parse)          //Convert to integers
                .Select(            
                    (x,n)=>             //Lambda with x being the part of the version and n being the index in the collection
                        n==i                    
                            ?x+1        //If n is the index to update increment x
                            :n>i        //Else if n is greater than index to update
                                ?0      //Set to zero
                                :x));   //Otherwise return x

Try it here

Python 2, 84 Bytes

I feel like this could really be shorter.. Might need a way to have a non-enumerate option.

lambda s,n:'.'.join(str([-~int(x)*(i==n),x][i<n])for i,x in enumerate(s.split('.')))

If we were able to take the version as a list of strings, there's a 75-byte solution:

g=lambda s,n:(str(-~int(s[0]))+'.0'*~-len(s))*(n<1)or s[0]+'.'+g(s[1:],n-1)

Furthermore, if both the input and output were lists of numbers, there's a 64-byte solution:

g=lambda s,n:([-~s[0]]+[0]*~-len(s))*(n<1)or [s[0]]+g(s[1:],n-1)

V 14 20 bytes

Again, I had to add code for the corner case of incrementing the final digit. (1-indexed)

DJA.0@"t.ò2wcw0òx

TryItOnline

Unprintables:

DJA.^[0@"t.^Aò2wcw0^[òx

This takes the arguments in reverse order, as separate lines:

3
1.2.3.4.5

Batch, 119 bytes

@set s=%1
@set i=%2
@set t=
@for %%i in (%s:.=,%)do @set/an=!!i*(%%i+!(i-=!!i))&call set t=%%t%%.%%n%%
@echo %t:~1%

1-indexed.

Perl 6, 67 bytes, 0-indexed

->$v,$i {$v.split('.').map({++$>$i??($++??0!!$_+1)!!$_}).join: '.'}

Explanation:

->$v,$i {$v.split('.').map({++$>$i??($++??0!!$_+1)!!$_}).join: '.'}
->$v,$i {                                                         } # function taking version/index to increment
         $v.split('.')                                              # split by dot
                      .map({                          })            # for each version number
                            ++$>$i??                                # if an anonymous variable ($), incremented,
                                                                    #  is greater than $i (index to increment)
                                    ($++??       )                  # if it's not the first time we've been over $i
                                                                    # (using another anonymous value, which gets default-init'd to 0)
                                          0                         # then 0 (reset lower version numbers)
                                           !!$_+1                   # otherwise, increment the number at $i
                                                  !!$_              # otherwise return the number part
                                                        .join: '.'  # join with a dot

PowerShell 3+, 75 74 bytes

($args[0]-split'\.'|%{$m=!$b;$b=$b-or$args[1]-eq$i++;(+$b+$_)*$m})-join'.'

Ungolfed

(
    $args[0] -split '\.' | 
        ForEach-Object -Process {
            $m= -not $b
            $b = $b -or ($args[1] -eq $i++)
            (([int]$b) + $_) * $m
        }
) -join '.'

Explanation

Parameters are accepted using the $args array.

1. Split the version string on ., then for each element:
   1. $m is set to be -not $b. On first run, $b will be undefined which will be coalesced to $false, so $m will start as $true. $m is intended to be a multiplier that's always 0 or 1 and it will be used later. $m must be evaluated here because we want it to be based on the last iteration's $b value.
   2. $b is set to itself -or the result of comparing an iterator $i with $args[1] (the index parameter). This means $b will be set to $true here once we're on the element that is to be incremented. Additionally, it will be $true in every subsequent iteration because the conditional is -or'd with its current value.
   3. $b is converted to a number using unary + ($false => 0, $true => 1), then added to the current version element $_ which is a [string], but PowerShell always tries to coalesce the argument on the right to the type on the left, so arithmetic will be performed, not string concatenation. Then this value will be multiplied by $m, which is still [bool] but will be implicitly coalesced.
2. Re-join the resulting array with ..

So, the first iteration where $b becomes $true, $b would have been $false when $m was evaluated, making $m equal $true, which will keep the multiplier at 1.

During that run $b becomes $true and is added to the version element (as 1), thereby incrementing it, and since the multiplier is still 1, that will be the end result.

So on the next iteration, $b will already be $true, making $m equal $false, which will make the multiplier 0. Since $b will forever be $true now, the multiplier will always be 0, so every element returned will be 0 too.

R, 100 95 92 86 bytes

Unusually for R, this uses 0-indexing. Anonymous function with two arguments (a string and an integer). Likely can be golfed down a tad.

function(v,i)cat((x=scan(t=el(strsplit(v,"\\."))))+c(rep(0,i),1,-x[-(0:i+1)]),sep=".")

05AB1E, 22 bytes

'.¡vy²N‹i0*}²NQi>})'.ý

Try it online!

I don't know how to do if-else in 05AB1E, so this is longer than it should be.

Coffee-script: 77 67 Bytes

f=(p,i)->((z<i&&v||z==i&&~~v+1||0)for v,z in p.split '.').join '.'

Woot! Time for cake and coffee for the beta release.

Thanks to @ven and @Cyoce I shaved 10 Bytes!

Python 3, 89 86 bytes

lambda v,c:'.'.join((str((int(x)+1)*(i==c)),x)[i<c]for i,x in enumerate(v.split('.')))

very naive way of getting things done

Edit: rewritten the conditional by referring to @kade

PHP, 81 bytes

awfully long. At least: the Elephpant still beats the Python.

foreach(explode(".",$argv[1])as$i=>$v)echo"."[!$i],($i<=$n=$argv[2])*$v+($i==$n);

loops through first argument split by dots: "."[!$i] is empty for the first and a dot for every other element; ($i<=$n) and ($i==$n) are implicitly cast to integer 0 or 1 for integer arithmetics.

JavaScript (ES6), 57 55 bytes

(s,x)=>s.split`.`.map((j,i)=>i==x?+j+1:i>x?0:j).join`.`

Examples:

n=(s,x)=>s.split`.`.map((j,i)=>i==x?+j+1:i>x?0:j).join`.`

console.log(n('1.0.3', 0))
console.log(n('1.2.3.4.5', 2))
console.log(n('10.0', 0))
console.log(n('3', 0))
console.log(n('1', 7))

Not the best JS implementation but it's fairly simple and follows the logic you'd expect.

Pyth - 21 bytes

j\.++<Kcz\.Qhs@KQmZ>K

Test Suite

Powershell, 80 100 95 92 Bytes

Saved 5 bytes by using a const for the -1..if

Saved 3 bytes by using !$b instead of $b-eq0

filter x($a,$b){[int[]]$y=$a.Split('.');-1..((-$b,-1)[!$b])|%{$y[$_]=0};$y[$b]++;$y-join'.'}

Explanation:

filter x($a,$b){
    [int[]]$y=$a.Split('.') #Split input into integer array
    $y[$b]++ #Increment 'major' version no. ($b) by one
    -1..((-$b,-1)[!$b])|%{$y[$_]=0} #Set all trailing numbers to 0, now also checks for $b=0 cases.
    $y-join'.' #Join back into '.' seperated Array
}

Test Cases:

x "1.0.3" 0
x "1.2.3.4.5" 2
x "10.0" 0
x "1" 7
2.0.0
1.2.4.0.0
11.0
Index was outside the bounds of the array.

Objective-C 531 Bytes

#import<Foundation/Foundation.h>
int main(int argc,const char *argv[]){@autoreleasepool{NSString *s=[NSString stringWithUTF8String:argv[1]];NSInteger n=strtol(argv[2],NULL,0);NSArray *c=[s componentsSeparatedByString:@"."];if(c.count<=n)NSLog(@"ERROR");else{int i=0;NSMutableString *v=[[NSMutableString alloc]init];for(;i<n;++i)[v appendFormat:@"%@.",[c objectAtIndex:i]];[v appendFormat:@"%li", strtol(((NSString *)[c objectAtIndex:i++]).UTF8String,NULL,0)+1l];for(;i<c.count;++i)[v appendString:@".0"];NSLog(@"%@",v);}}return 0;}

compile:

clang -fobjc-arc -Os main.m -o main

usage:

./main 1.2.3 1

Javascript ES6: 60 bytes

n.split(".").map((n,r)=>{return r>i?n=0:n}).join("."),n[i]++}

R, 75 bytes

f=function(a,b){n=scan(t=a,se=".");m=-n;m[b]=1;m[1:b-1]=0;cat(n+m,sep=".")}

Indexing is 1-based. You can play with it online here.

APL (Dyalog), 31 bytes

Requires ⎕IO←0 (Index Origin 0), which is default on many systems. Full program body; prompts for text input (version) and then numeric input (index).

' '⎕R'.'⍕⎕(⊢∘≢↑↑,1+⊃)⊃⌽'.'⎕VFI⍞

Try it online!

⍞ prompt for text input

'.'⎕VFI Verify and Fix Input using period as field separator (fields' validities, fields' values)

⌽ reverse (to put the values in front)

⊃ pick the first (the values)

⎕(...) apply the following tacit function to that, using evaluated input as left argument:

_{To explain the non-tacit equivalents of each function application we will now use ⍺ to indicate the left argument (the index) and ⍵ to indicate the right argument (the list of individual numbers of the originally inputted current version number).}

 ⊃ equivalent to (⍺⊃⍵) use ⍺ to pick an element from ⍵

 1+ add one to that 

 ↑, equivalent to (⍺↑⍵), prepend ⍺ numbers taken from ⍵

 ⊢∘≢↑ equivalent to (⍺⊢∘≢⍵)↑ equivalent to (≢⍵)↑ take as many numbers from that as there are elements in ⍵, padding with zeros if necessary

⍕ format (stringify with one space between each number)

' '⎕R'.' PCRE Replace spaces with periods

Java 8, 130 bytes

s->n->{String r="",a[]=s.split("\\.");for(int l=a.length,i=-1;++i<l;)r+=(i>n?0:new Long(a[i])+(i<n?0:1))+(i<l-1?".":"");return r;}

Explanation:

Try it here.

s->n->{                 // Method with String and Integer parameters and String return-type
  String r="",          //  Result-String
  a[]=s.split("\\.");   //  String-array split by the dots
  for(int l=a.length,   //  Length of the array
      i=-1;++i<l;)      //  Loop from 0 to `l` (exclusive)
    r+=                 //   Append the result-String with:
       (i>n?            //    If index `i` is beyond input `n`:
        0               //     Append a zero
       :                //    Else:
        new Long(a[i])  //     Convert the current String to a number
        +(i<n?          //     If index `i` is before input `n`
           0            //      Leave the number the same by adding 0
          :             //     Else:
           1))          //      Add 1 to raise the version at index `n`
       +(i<l-1?         //    If we've haven't reached the last iteration yet:
          "."           //     Append a dot
         :              //    Else:
          ""            //     Append nothing
   );                   //  End of loop
   return r;            //  Return the result-String
}                       // End of method

LiveScript, 53 52 bytes

->(for e,i in it/\.
 [+e+1;0;e][(i>&1)+2*(i<&1)])*\.

-1 byte thanks to @ASCII-only!

Old Explanation:

(a,b)->x=a/\.;x[b]++;(x[to b] ++ [0]*(x.length-1-b))*\.
(a,b)->                                                 # a function taking a and b (version and index)
       x=a/\.;                                          # split a on dot, store in x
              x[b]++;                                   # increment at the given index
                     (x[to b]                           # slice be from 0 to the index
                              ++                        # concat (both spaces are necessary so it's not interpreted as an increment operator
                                 [0]*(x.length-1-b))    # with enough zeros to fill the array back to its original size (x's size)
                                                    *\. # join on dot

Another self-answer... Not that anyone golfes in LiveScript anyway. :P

I was working on another version:

(a,b)->(a/\.=>..[b]++;..[b to *]=0)*\.

But * is too overloaded to be recognized in a splicing index, thus =0 will try to access 0[0]. So you need to write something like ..[b to ..length- b]=[0]*(..length-1-b) and it's longer in the end.

Haskell, 136 129 bytes

import Data.List.Split
import Data.List
f s n|(x,z:y)<-splitAt n(splitOn "."s)=intercalate "."(x++[show$read z+1]++map(\_->"0")y)

Try it online!

Original