Decimal Digit Truncating

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite












Input:



A decimal



Output:



The input truncated



How to truncate them?



The integer part of the input will remain the same. The decimal part of the input will be truncated based on the lowest digit in consecutive order. The two step-by-step examples below should clarify what I mean by that:



Example 1: Input: 17.46940822



Step 1: Leave the integer part unchanged: 17

Step 2: Look at the first decimal digit, and truncate it to that many decimal digits: 4 → .4694

Step 3: Look at the next decimal digit. If it's lower, adjust the truncation accordingly. If not, continue to the next decimal digit: 6 → remains .4694

Step 3 again: 9 → remains .4694

Step 3 again: 4 → remains .4694



So the output will be: 17.4694.



Example 2: Input: 10.4521084



Step 1: Leave the integer part unchanged: 10

Step 2: Loop at the first decimal digit, and truncate it to that many decimal digits: 4 → .4521

Step 3: Look at the next decimal digit. If it's lower, adjust the truncation accordingly. If not, continue to the next decimal digit: 5 → remains .4521

Step 3 again: 2 → truncation adjusted to: .45



So the output will be: 10.45



Challenge rules:



  • You are allowed to take the input as decimal, string, list/array/stream of characters, etc. (Same applies to the output, although I doubt there are a lot of languages that have the proper truncation while remain working with decimal values.)

    You are also allowed to use a comma , instead of a dot . (perhaps useful for regex-based solution/languages).

  • If the decimal part is not long enough for the truncation, you are optionally allowed to add trailing zeros. I.e. 123.4 can be both 123.4 or 123.4000

  • If you have to truncate to 0 decimal digits we omit the decimal digits. The dot/comma may remain optionally if you use strings. I.e. 847.3072 → 847 (or 847.)

  • You can assume the input will always contain at least one decimal digit, so 123 won't be an input, but 123.0 can be an input.

General rules:



  • This is code-golf, so shortest answer in bytes wins.

    Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.


  • Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.


  • Default Loopholes are forbidden.

  • If possible, please add a link with a test for your code (i.e. TIO).

  • Also, adding an explanation for your answer is highly recommended.

Test cases:



Input: Output:

17.46940822 17.4694
10.4521084 10.45
123.4 123.4000 (or 123.4)
-847.3072 -847 (or -847.)
123.0 123 (or 123.)
98.76543210 98.7654
0.3165 0.3
-0.1365 -0.1
82937.82937 82937.82
-0.1010101 -0.1
-1.0101010 -1 (or -1.)
28.684297 28.68
13.35 13.350 (or 13.35)





code-golf number floating-point







share|improve this question























  • Need we output string?
    – l4m2
    2 hours ago










  • @l4m2 No, you are also allowed to output it as a decimal, or list of characters. Although I don't know a lot of languages that can do the proper truncation without converting to string format.
    – Kevin Cruijssen
    2 hours ago










  • So it's fine to remove ending zero if outputing float?
    – l4m2
    2 hours ago










  • @l4m2 Sure. I've updated the rules to also allow no additional trailing zeros.
    – Kevin Cruijssen
    2 hours ago














up vote
1
down vote

favorite












Input:



A decimal



Output:



The input truncated



How to truncate them?



The integer part of the input will remain the same. The decimal part of the input will be truncated based on the lowest digit in consecutive order. The two step-by-step examples below should clarify what I mean by that:



Example 1: Input: 17.46940822



Step 1: Leave the integer part unchanged: 17

Step 2: Look at the first decimal digit, and truncate it to that many decimal digits: 4 → .4694

Step 3: Look at the next decimal digit. If it's lower, adjust the truncation accordingly. If not, continue to the next decimal digit: 6 → remains .4694

Step 3 again: 9 → remains .4694

Step 3 again: 4 → remains .4694



So the output will be: 17.4694.



Example 2: Input: 10.4521084



Step 1: Leave the integer part unchanged: 10

Step 2: Loop at the first decimal digit, and truncate it to that many decimal digits: 4 → .4521

Step 3: Look at the next decimal digit. If it's lower, adjust the truncation accordingly. If not, continue to the next decimal digit: 5 → remains .4521

Step 3 again: 2 → truncation adjusted to: .45



So the output will be: 10.45



Challenge rules:



  • You are allowed to take the input as decimal, string, list/array/stream of characters, etc. (Same applies to the output, although I doubt there are a lot of languages that have the proper truncation while remain working with decimal values.)

    You are also allowed to use a comma , instead of a dot . (perhaps useful for regex-based solution/languages).

  • If the decimal part is not long enough for the truncation, you are optionally allowed to add trailing zeros. I.e. 123.4 can be both 123.4 or 123.4000

  • If you have to truncate to 0 decimal digits we omit the decimal digits. The dot/comma may remain optionally if you use strings. I.e. 847.3072 → 847 (or 847.)

  • You can assume the input will always contain at least one decimal digit, so 123 won't be an input, but 123.0 can be an input.

General rules:



  • This is code-golf, so shortest answer in bytes wins.

    Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.


  • Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.


  • Default Loopholes are forbidden.

  • If possible, please add a link with a test for your code (i.e. TIO).

  • Also, adding an explanation for your answer is highly recommended.

Test cases:



Input: Output:

17.46940822 17.4694
10.4521084 10.45
123.4 123.4000 (or 123.4)
-847.3072 -847 (or -847.)
123.0 123 (or 123.)
98.76543210 98.7654
0.3165 0.3
-0.1365 -0.1
82937.82937 82937.82
-0.1010101 -0.1
-1.0101010 -1 (or -1.)
28.684297 28.68
13.35 13.350 (or 13.35)





code-golf number floating-point







share|improve this question























  • Need we output string?
    – l4m2
    2 hours ago










  • @l4m2 No, you are also allowed to output it as a decimal, or list of characters. Although I don't know a lot of languages that can do the proper truncation without converting to string format.
    – Kevin Cruijssen
    2 hours ago










  • So it's fine to remove ending zero if outputing float?
    – l4m2
    2 hours ago










  • @l4m2 Sure. I've updated the rules to also allow no additional trailing zeros.
    – Kevin Cruijssen
    2 hours ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Input:



A decimal



Output:



The input truncated



How to truncate them?



The integer part of the input will remain the same. The decimal part of the input will be truncated based on the lowest digit in consecutive order. The two step-by-step examples below should clarify what I mean by that:



Example 1: Input: 17.46940822



Step 1: Leave the integer part unchanged: 17

Step 2: Look at the first decimal digit, and truncate it to that many decimal digits: 4 → .4694

Step 3: Look at the next decimal digit. If it's lower, adjust the truncation accordingly. If not, continue to the next decimal digit: 6 → remains .4694

Step 3 again: 9 → remains .4694

Step 3 again: 4 → remains .4694



So the output will be: 17.4694.



Example 2: Input: 10.4521084



Step 1: Leave the integer part unchanged: 10

Step 2: Loop at the first decimal digit, and truncate it to that many decimal digits: 4 → .4521

Step 3: Look at the next decimal digit. If it's lower, adjust the truncation accordingly. If not, continue to the next decimal digit: 5 → remains .4521

Step 3 again: 2 → truncation adjusted to: .45



So the output will be: 10.45



Challenge rules:



  • You are allowed to take the input as decimal, string, list/array/stream of characters, etc. (Same applies to the output, although I doubt there are a lot of languages that have the proper truncation while remain working with decimal values.)

    You are also allowed to use a comma , instead of a dot . (perhaps useful for regex-based solution/languages).

  • If the decimal part is not long enough for the truncation, you are optionally allowed to add trailing zeros. I.e. 123.4 can be both 123.4 or 123.4000

  • If you have to truncate to 0 decimal digits we omit the decimal digits. The dot/comma may remain optionally if you use strings. I.e. 847.3072 → 847 (or 847.)

  • You can assume the input will always contain at least one decimal digit, so 123 won't be an input, but 123.0 can be an input.

General rules:



  • This is code-golf, so shortest answer in bytes wins.

    Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.


  • Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.


  • Default Loopholes are forbidden.

  • If possible, please add a link with a test for your code (i.e. TIO).

  • Also, adding an explanation for your answer is highly recommended.

Test cases:



Input: Output:

17.46940822 17.4694
10.4521084 10.45
123.4 123.4000 (or 123.4)
-847.3072 -847 (or -847.)
123.0 123 (or 123.)
98.76543210 98.7654
0.3165 0.3
-0.1365 -0.1
82937.82937 82937.82
-0.1010101 -0.1
-1.0101010 -1 (or -1.)
28.684297 28.68
13.35 13.350 (or 13.35)





code-golf number floating-point







share|improve this question















Input:



A decimal



Output:



The input truncated



How to truncate them?



The integer part of the input will remain the same. The decimal part of the input will be truncated based on the lowest digit in consecutive order. The two step-by-step examples below should clarify what I mean by that:



Example 1: Input: 17.46940822



Step 1: Leave the integer part unchanged: 17

Step 2: Look at the first decimal digit, and truncate it to that many decimal digits: 4 → .4694

Step 3: Look at the next decimal digit. If it's lower, adjust the truncation accordingly. If not, continue to the next decimal digit: 6 → remains .4694

Step 3 again: 9 → remains .4694

Step 3 again: 4 → remains .4694



So the output will be: 17.4694.



Example 2: Input: 10.4521084



Step 1: Leave the integer part unchanged: 10

Step 2: Loop at the first decimal digit, and truncate it to that many decimal digits: 4 → .4521

Step 3: Look at the next decimal digit. If it's lower, adjust the truncation accordingly. If not, continue to the next decimal digit: 5 → remains .4521

Step 3 again: 2 → truncation adjusted to: .45



So the output will be: 10.45



Challenge rules:



  • You are allowed to take the input as decimal, string, list/array/stream of characters, etc. (Same applies to the output, although I doubt there are a lot of languages that have the proper truncation while remain working with decimal values.)

    You are also allowed to use a comma , instead of a dot . (perhaps useful for regex-based solution/languages).

  • If the decimal part is not long enough for the truncation, you are optionally allowed to add trailing zeros. I.e. 123.4 can be both 123.4 or 123.4000

  • If you have to truncate to 0 decimal digits we omit the decimal digits. The dot/comma may remain optionally if you use strings. I.e. 847.3072 → 847 (or 847.)

  • You can assume the input will always contain at least one decimal digit, so 123 won't be an input, but 123.0 can be an input.

General rules:



  • This is code-golf, so shortest answer in bytes wins.

    Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.


  • Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.


  • Default Loopholes are forbidden.

  • If possible, please add a link with a test for your code (i.e. TIO).

  • Also, adding an explanation for your answer is highly recommended.

Test cases:



Input: Output:

17.46940822 17.4694
10.4521084 10.45
123.4 123.4000 (or 123.4)
-847.3072 -847 (or -847.)
123.0 123 (or 123.)
98.76543210 98.7654
0.3165 0.3
-0.1365 -0.1
82937.82937 82937.82
-0.1010101 -0.1
-1.0101010 -1 (or -1.)
28.684297 28.68
13.35 13.350 (or 13.35)





code-golf number floating-point




code-golf number floating-point






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 2 hours ago

























asked 2 hours ago









Kevin Cruijssen

32.6k554176




32.6k554176











  • Need we output string?
    – l4m2
    2 hours ago










  • @l4m2 No, you are also allowed to output it as a decimal, or list of characters. Although I don't know a lot of languages that can do the proper truncation without converting to string format.
    – Kevin Cruijssen
    2 hours ago










  • So it's fine to remove ending zero if outputing float?
    – l4m2
    2 hours ago










  • @l4m2 Sure. I've updated the rules to also allow no additional trailing zeros.
    – Kevin Cruijssen
    2 hours ago
















  • Need we output string?
    – l4m2
    2 hours ago










  • @l4m2 No, you are also allowed to output it as a decimal, or list of characters. Although I don't know a lot of languages that can do the proper truncation without converting to string format.
    – Kevin Cruijssen
    2 hours ago










  • So it's fine to remove ending zero if outputing float?
    – l4m2
    2 hours ago










  • @l4m2 Sure. I've updated the rules to also allow no additional trailing zeros.
    – Kevin Cruijssen
    2 hours ago















Need we output string?
– l4m2
2 hours ago




Need we output string?
– l4m2
2 hours ago












@l4m2 No, you are also allowed to output it as a decimal, or list of characters. Although I don't know a lot of languages that can do the proper truncation without converting to string format.
– Kevin Cruijssen
2 hours ago




@l4m2 No, you are also allowed to output it as a decimal, or list of characters. Although I don't know a lot of languages that can do the proper truncation without converting to string format.
– Kevin Cruijssen
2 hours ago












So it's fine to remove ending zero if outputing float?
– l4m2
2 hours ago




So it's fine to remove ending zero if outputing float?
– l4m2
2 hours ago












@l4m2 Sure. I've updated the rules to also allow no additional trailing zeros.
– Kevin Cruijssen
2 hours ago




@l4m2 Sure. I've updated the rules to also allow no additional trailing zeros.
– Kevin Cruijssen
2 hours ago










4 Answers
4






active

oldest

votes

















up vote
1
down vote














Python 2, 103 bytes





def f(n):
 a=n.find('.');i,w=1,9
 while i<=w and a+i<len(n):w=min(int(n[a+i]),w);i+=1
 return n[:a+w+1]


Try it online!






share|improve this answer




















  • a+w+1 can be a-~w. Nice answer though, +1 from me.
    – Kevin Cruijssen
    1 hour ago






  • 1




    Oh, and return can be print as well. And here you are golfing Python for years. Tsk tsk tsk. ;p (It's the simplest things we usually forgot to golf I guess.)
    – Kevin Cruijssen
    1 hour ago











  • Instead of a function, use n=input() and remove the spaces and print as mentioned by @KevinCruijssen to save a few bytes.
    – Vedant Kandoi
    1 hour ago


















up vote
1
down vote














Charcoal, 28 bytes



≔⪪Ｓ.θＷ⊟ＥθΦκ‹ＩμＬκ⊞θ…⊟θＩ§ι⁰⪫θ.


Try it online! Link is to verbose version of code. Explanation:



≔⪪Ｓ.θ


Split the input on ..



Ｗ⊟ＥθΦκ‹ＩμＬκ


Filter each part on those digits that are less than the string's length. Then take the last result. (This saves 1 byte over filtering on just the decimals.) Loop until that is empty.



⊞θ…⊟θＩ§ι⁰


Take the first filtered digit and truncate the decimals to that length.



⪫θ.


Join the parts back together and implicitly print.






share|improve this answer




















  • Seems to give an incorrect result for the negative input-cases (although it is possible I don't know how to input negative values in Charcoal, I tried with a leading - or _, but both failed). Everything else works fine.
    – Kevin Cruijssen
    31 mins ago

















up vote
1
down vote














05AB1E, 20 bytes



4 bytes extra due to a bug with £ not handling 0 correctly.



'.¡`DvÐgsNè-F¨]'.ý0Ü


Try it online!
or as a Test Suite






share|improve this answer




















  • Where is the 0Ãœ used for? It seems to give the same results without removing trailing zeros. Did you try a test case I should add to the challenge description for which the 0Ãœ is required?
    – Kevin Cruijssen
    26 mins ago


















up vote
0
down vote














C (gcc), 84 bytes



Takes input as a pointer to a string buffer, outputs by modification. Note that providing a pointer to a buffer that is too small (intPartLen + 1 + 9 > bufSize) can cause SEGFAULT.



If assuming the size of the buffer to be sufficient is not allowed, +7 to the score.





l,m,M;f(char*F)F=strchr(F,46);M=10;for(l=0;F[++l]&&l<M;m<M?M=m:0)m=F[l]-47;F[M]=0;


Degolf



l,m,M;f(char*F)
 F=strchr(F,46); // Find the pointer to the period
 M=10;for(l=0; // Initialize variables for the loop.
 F[++l]&&l<M; // Check for null terminator, increment length, and check
 // that we're not checking beyond limits of the recursion.
 m<M?M=m:0) // Keep smaller of M and m
 m=F[l]-47; // Read the character from F, deduct 47 to map 0->1, 9->10
 F[M]=0; // Put a null terminator at the index M.



Try it online!





share




















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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote














    Python 2, 103 bytes





    def f(n):
     a=n.find('.');i,w=1,9
     while i<=w and a+i<len(n):w=min(int(n[a+i]),w);i+=1
     return n[:a+w+1]


    Try it online!






    share|improve this answer




















    • a+w+1 can be a-~w. Nice answer though, +1 from me.
      – Kevin Cruijssen
      1 hour ago






    • 1




      Oh, and return can be print as well. And here you are golfing Python for years. Tsk tsk tsk. ;p (It's the simplest things we usually forgot to golf I guess.)
      – Kevin Cruijssen
      1 hour ago











    • Instead of a function, use n=input() and remove the spaces and print as mentioned by @KevinCruijssen to save a few bytes.
      – Vedant Kandoi
      1 hour ago















    up vote
    1
    down vote














    Python 2, 103 bytes





    def f(n):
     a=n.find('.');i,w=1,9
     while i<=w and a+i<len(n):w=min(int(n[a+i]),w);i+=1
     return n[:a+w+1]


    Try it online!






    share|improve this answer




















    • a+w+1 can be a-~w. Nice answer though, +1 from me.
      – Kevin Cruijssen
      1 hour ago






    • 1




      Oh, and return can be print as well. And here you are golfing Python for years. Tsk tsk tsk. ;p (It's the simplest things we usually forgot to golf I guess.)
      – Kevin Cruijssen
      1 hour ago











    • Instead of a function, use n=input() and remove the spaces and print as mentioned by @KevinCruijssen to save a few bytes.
      – Vedant Kandoi
      1 hour ago













    up vote
    1
    down vote










    up vote
    1
    down vote










    Python 2, 103 bytes





    def f(n):
     a=n.find('.');i,w=1,9
     while i<=w and a+i<len(n):w=min(int(n[a+i]),w);i+=1
     return n[:a+w+1]


    Try it online!






    share|improve this answer













    Python 2, 103 bytes





    def f(n):
     a=n.find('.');i,w=1,9
     while i<=w and a+i<len(n):w=min(int(n[a+i]),w);i+=1
     return n[:a+w+1]


    Try it online!







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 2 hours ago









    TFeld

    12.7k2836




    12.7k2836











    • a+w+1 can be a-~w. Nice answer though, +1 from me.
      – Kevin Cruijssen
      1 hour ago






    • 1




      Oh, and return can be print as well. And here you are golfing Python for years. Tsk tsk tsk. ;p (It's the simplest things we usually forgot to golf I guess.)
      – Kevin Cruijssen
      1 hour ago











    • Instead of a function, use n=input() and remove the spaces and print as mentioned by @KevinCruijssen to save a few bytes.
      – Vedant Kandoi
      1 hour ago

















    • a+w+1 can be a-~w. Nice answer though, +1 from me.
      – Kevin Cruijssen
      1 hour ago






    • 1




      Oh, and return can be print as well. And here you are golfing Python for years. Tsk tsk tsk. ;p (It's the simplest things we usually forgot to golf I guess.)
      – Kevin Cruijssen
      1 hour ago











    • Instead of a function, use n=input() and remove the spaces and print as mentioned by @KevinCruijssen to save a few bytes.
      – Vedant Kandoi
      1 hour ago
















    a+w+1 can be a-~w. Nice answer though, +1 from me.
    – Kevin Cruijssen
    1 hour ago




    a+w+1 can be a-~w. Nice answer though, +1 from me.
    – Kevin Cruijssen
    1 hour ago




    1




    1




    Oh, and return can be print as well. And here you are golfing Python for years. Tsk tsk tsk. ;p (It's the simplest things we usually forgot to golf I guess.)
    – Kevin Cruijssen
    1 hour ago





    Oh, and return can be print as well. And here you are golfing Python for years. Tsk tsk tsk. ;p (It's the simplest things we usually forgot to golf I guess.)
    – Kevin Cruijssen
    1 hour ago













    Instead of a function, use n=input() and remove the spaces and print as mentioned by @KevinCruijssen to save a few bytes.
    – Vedant Kandoi
    1 hour ago





    Instead of a function, use n=input() and remove the spaces and print as mentioned by @KevinCruijssen to save a few bytes.
    – Vedant Kandoi
    1 hour ago











    up vote
    1
    down vote














    Charcoal, 28 bytes



    ≔⪪Ｓ.θＷ⊟ＥθΦκ‹ＩμＬκ⊞θ…⊟θＩ§ι⁰⪫θ.


    Try it online! Link is to verbose version of code. Explanation:



    ≔⪪Ｓ.θ


    Split the input on ..



    Ｗ⊟ＥθΦκ‹ＩμＬκ


    Filter each part on those digits that are less than the string's length. Then take the last result. (This saves 1 byte over filtering on just the decimals.) Loop until that is empty.



    ⊞θ…⊟θＩ§ι⁰


    Take the first filtered digit and truncate the decimals to that length.



    ⪫θ.


    Join the parts back together and implicitly print.






    share|improve this answer




















    • Seems to give an incorrect result for the negative input-cases (although it is possible I don't know how to input negative values in Charcoal, I tried with a leading - or _, but both failed). Everything else works fine.
      – Kevin Cruijssen
      31 mins ago














    up vote
    1
    down vote














    Charcoal, 28 bytes



    ≔⪪Ｓ.θＷ⊟ＥθΦκ‹ＩμＬκ⊞θ…⊟θＩ§ι⁰⪫θ.


    Try it online! Link is to verbose version of code. Explanation:



    ≔⪪Ｓ.θ


    Split the input on ..



    Ｗ⊟ＥθΦκ‹ＩμＬκ


    Filter each part on those digits that are less than the string's length. Then take the last result. (This saves 1 byte over filtering on just the decimals.) Loop until that is empty.



    ⊞θ…⊟θＩ§ι⁰


    Take the first filtered digit and truncate the decimals to that length.



    ⪫θ.


    Join the parts back together and implicitly print.






    share|improve this answer




















    • Seems to give an incorrect result for the negative input-cases (although it is possible I don't know how to input negative values in Charcoal, I tried with a leading - or _, but both failed). Everything else works fine.
      – Kevin Cruijssen
      31 mins ago












    up vote
    1
    down vote










    up vote
    1
    down vote










    Charcoal, 28 bytes



    ≔⪪Ｓ.θＷ⊟ＥθΦκ‹ＩμＬκ⊞θ…⊟θＩ§ι⁰⪫θ.


    Try it online! Link is to verbose version of code. Explanation:



    ≔⪪Ｓ.θ


    Split the input on ..



    Ｗ⊟ＥθΦκ‹ＩμＬκ


    Filter each part on those digits that are less than the string's length. Then take the last result. (This saves 1 byte over filtering on just the decimals.) Loop until that is empty.



    ⊞θ…⊟θＩ§ι⁰


    Take the first filtered digit and truncate the decimals to that length.



    ⪫θ.


    Join the parts back together and implicitly print.






    share|improve this answer













    Charcoal, 28 bytes



    ≔⪪Ｓ.θＷ⊟ＥθΦκ‹ＩμＬκ⊞θ…⊟θＩ§ι⁰⪫θ.


    Try it online! Link is to verbose version of code. Explanation:



    ≔⪪Ｓ.θ


    Split the input on ..



    Ｗ⊟ＥθΦκ‹ＩμＬκ


    Filter each part on those digits that are less than the string's length. Then take the last result. (This saves 1 byte over filtering on just the decimals.) Loop until that is empty.



    ⊞θ…⊟θＩ§ι⁰


    Take the first filtered digit and truncate the decimals to that length.



    ⪫θ.


    Join the parts back together and implicitly print.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 36 mins ago









    Neil

    77.1k744174




    77.1k744174











    • Seems to give an incorrect result for the negative input-cases (although it is possible I don't know how to input negative values in Charcoal, I tried with a leading - or _, but both failed). Everything else works fine.
      – Kevin Cruijssen
      31 mins ago
















    • Seems to give an incorrect result for the negative input-cases (although it is possible I don't know how to input negative values in Charcoal, I tried with a leading - or _, but both failed). Everything else works fine.
      – Kevin Cruijssen
      31 mins ago















    Seems to give an incorrect result for the negative input-cases (although it is possible I don't know how to input negative values in Charcoal, I tried with a leading - or _, but both failed). Everything else works fine.
    – Kevin Cruijssen
    31 mins ago




    Seems to give an incorrect result for the negative input-cases (although it is possible I don't know how to input negative values in Charcoal, I tried with a leading - or _, but both failed). Everything else works fine.
    – Kevin Cruijssen
    31 mins ago










    up vote
    1
    down vote














    05AB1E, 20 bytes



    4 bytes extra due to a bug with £ not handling 0 correctly.



    '.¡`DvÐgsNè-F¨]'.ý0Ü


    Try it online!
    or as a Test Suite






    share|improve this answer




















    • Where is the 0Ãœ used for? It seems to give the same results without removing trailing zeros. Did you try a test case I should add to the challenge description for which the 0Ãœ is required?
      – Kevin Cruijssen
      26 mins ago















    up vote
    1
    down vote














    05AB1E, 20 bytes



    4 bytes extra due to a bug with £ not handling 0 correctly.



    '.¡`DvÐgsNè-F¨]'.ý0Ü


    Try it online!
    or as a Test Suite






    share|improve this answer




















    • Where is the 0Ãœ used for? It seems to give the same results without removing trailing zeros. Did you try a test case I should add to the challenge description for which the 0Ãœ is required?
      – Kevin Cruijssen
      26 mins ago













    up vote
    1
    down vote










    up vote
    1
    down vote










    05AB1E, 20 bytes



    4 bytes extra due to a bug with £ not handling 0 correctly.



    '.¡`DvÐgsNè-F¨]'.ý0Ü


    Try it online!
    or as a Test Suite






    share|improve this answer













    05AB1E, 20 bytes



    4 bytes extra due to a bug with £ not handling 0 correctly.



    '.¡`DvÐgsNè-F¨]'.ý0Ü


    Try it online!
    or as a Test Suite







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 29 mins ago









    Emigna

    44.1k431133




    44.1k431133











    • Where is the 0Ãœ used for? It seems to give the same results without removing trailing zeros. Did you try a test case I should add to the challenge description for which the 0Ãœ is required?
      – Kevin Cruijssen
      26 mins ago

















    • Where is the 0Ãœ used for? It seems to give the same results without removing trailing zeros. Did you try a test case I should add to the challenge description for which the 0Ãœ is required?
      – Kevin Cruijssen
      26 mins ago
















    Where is the 0Ü used for? It seems to give the same results without removing trailing zeros. Did you try a test case I should add to the challenge description for which the 0Ü is required?
    – Kevin Cruijssen
    26 mins ago





    Where is the 0Ü used for? It seems to give the same results without removing trailing zeros. Did you try a test case I should add to the challenge description for which the 0Ü is required?
    – Kevin Cruijssen
    26 mins ago











    up vote
    0
    down vote














    C (gcc), 84 bytes



    Takes input as a pointer to a string buffer, outputs by modification. Note that providing a pointer to a buffer that is too small (intPartLen + 1 + 9 > bufSize) can cause SEGFAULT.



    If assuming the size of the buffer to be sufficient is not allowed, +7 to the score.





    l,m,M;f(char*F)F=strchr(F,46);M=10;for(l=0;F[++l]&&l<M;m<M?M=m:0)m=F[l]-47;F[M]=0;


    Degolf



    l,m,M;f(char*F)
     F=strchr(F,46); // Find the pointer to the period
     M=10;for(l=0; // Initialize variables for the loop.
     F[++l]&&l<M; // Check for null terminator, increment length, and check
     // that we're not checking beyond limits of the recursion.
     m<M?M=m:0) // Keep smaller of M and m
     m=F[l]-47; // Read the character from F, deduct 47 to map 0->1, 9->10
     F[M]=0; // Put a null terminator at the index M.



    Try it online!





    share
























      up vote
      0
      down vote














      C (gcc), 84 bytes



      Takes input as a pointer to a string buffer, outputs by modification. Note that providing a pointer to a buffer that is too small (intPartLen + 1 + 9 > bufSize) can cause SEGFAULT.



      If assuming the size of the buffer to be sufficient is not allowed, +7 to the score.





      l,m,M;f(char*F)F=strchr(F,46);M=10;for(l=0;F[++l]&&l<M;m<M?M=m:0)m=F[l]-47;F[M]=0;


      Degolf



      l,m,M;f(char*F)
       F=strchr(F,46); // Find the pointer to the period
       M=10;for(l=0; // Initialize variables for the loop.
       F[++l]&&l<M; // Check for null terminator, increment length, and check
       // that we're not checking beyond limits of the recursion.
       m<M?M=m:0) // Keep smaller of M and m
       m=F[l]-47; // Read the character from F, deduct 47 to map 0->1, 9->10
       F[M]=0; // Put a null terminator at the index M.



      Try it online!





      share






















        up vote
        0
        down vote










        up vote
        0
        down vote










        C (gcc), 84 bytes



        Takes input as a pointer to a string buffer, outputs by modification. Note that providing a pointer to a buffer that is too small (intPartLen + 1 + 9 > bufSize) can cause SEGFAULT.



        If assuming the size of the buffer to be sufficient is not allowed, +7 to the score.





        l,m,M;f(char*F)F=strchr(F,46);M=10;for(l=0;F[++l]&&l<M;m<M?M=m:0)m=F[l]-47;F[M]=0;


        Degolf



        l,m,M;f(char*F)
         F=strchr(F,46); // Find the pointer to the period
         M=10;for(l=0; // Initialize variables for the loop.
         F[++l]&&l<M; // Check for null terminator, increment length, and check
         // that we're not checking beyond limits of the recursion.
         m<M?M=m:0) // Keep smaller of M and m
         m=F[l]-47; // Read the character from F, deduct 47 to map 0->1, 9->10
         F[M]=0; // Put a null terminator at the index M.



        Try it online!





        share













        C (gcc), 84 bytes



        Takes input as a pointer to a string buffer, outputs by modification. Note that providing a pointer to a buffer that is too small (intPartLen + 1 + 9 > bufSize) can cause SEGFAULT.



        If assuming the size of the buffer to be sufficient is not allowed, +7 to the score.





        l,m,M;f(char*F)F=strchr(F,46);M=10;for(l=0;F[++l]&&l<M;m<M?M=m:0)m=F[l]-47;F[M]=0;


        Degolf



        l,m,M;f(char*F)
         F=strchr(F,46); // Find the pointer to the period
         M=10;for(l=0; // Initialize variables for the loop.
         F[++l]&&l<M; // Check for null terminator, increment length, and check
         // that we're not checking beyond limits of the recursion.
         m<M?M=m:0) // Keep smaller of M and m
         m=F[l]-47; // Read the character from F, deduct 47 to map 0->1, 9->10
         F[M]=0; // Put a null terminator at the index M.



        Try it online!






        share











        share


        share










        answered 1 min ago









        Rogem

        61910




        61910



























             

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