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C, 30 + 1-4 byte f(m){f(printf("%0*i\n",m,0));} Has this limitations: counter stops working at INT_MAX or at stack overflow. Needs a input value to indicate on which line we start. 0 starts...

posted 1y ago by H_H‭  ·  edited 1y ago by H_H‭

Answer
#4: Post edited by user avatar H_H‭ · 2023-09-08T15:07:09Z (over 1 year ago)
  • ## C, 30 bytes + 1-4 byte
  • f(m){f(printf("%0*i\n",m,0));}
  • Has this limitations:
  • - counter stops working at `INT_MAX` or at stack overflow.
  • - Needs a input value to indicate on which line we start. `0` starts on the line with one character.
  • - It uses `0` and not `*` as output character
  • Because of the additional input number, i added the + 1 byte in the title. If you call the program, you need to write `f(0)`, this is why the +4 bytes in the title.
  • ### Fully working program, 36 bytes:
  • main(m){main(printf("%0*i\n",m,0));}
  • ## C, 30 + 1-4 byte
  • f(m){f(printf("%0*i\n",m,0));}
  • Has this limitations:
  • - counter stops working at `INT_MAX` or at stack overflow.
  • - Needs a input value to indicate on which line we start. `0` starts on the line with one character.
  • - It uses `0` and not `*` as output character
  • Because of the additional input number, i added the + 1 in the title. If you call the program, you need to write `f(0)`, this is why the +4 in the title.
  • ### Fully working program, 36 byte:
  • main(m){main(printf("%0*i\n",m,0));}
#3: Post edited by user avatar H_H‭ · 2023-09-04T16:39:09Z (over 1 year ago)
  • ## C, 30 bytes + 1-4 byte
  • `f(m){f(printf("%0*i
  • ",m,0));}`
  • Has this limitations:
  • - counter stops working at `INT_MAX` or at stack overflow.
  • - Needs a input value to indicate on which line we start. `0` starts on the line with one character.
  • - It uses `0` and not `*` as output character
  • Because of the additional input number, i added the + 1 byte in the title. If you call the program, you need to write `f(0)`, this is why the +4 bytes in the title.
  • ### Fully working program, 36 bytes:
  • `main(m){main(printf("%0*i
  • ",m,0));}`
  • ## C, 30 bytes + 1-4 byte
  • f(m){f(printf("%0*i
  • ",m,0));}
  • Has this limitations:
  • - counter stops working at `INT_MAX` or at stack overflow.
  • - Needs a input value to indicate on which line we start. `0` starts on the line with one character.
  • - It uses `0` and not `*` as output character
  • Because of the additional input number, i added the + 1 byte in the title. If you call the program, you need to write `f(0)`, this is why the +4 bytes in the title.
  • ### Fully working program, 36 bytes:
  • main(m){main(printf("%0*i
  • ",m,0));}
#2: Post edited by user avatar H_H‭ · 2023-08-30T14:17:07Z (over 1 year ago)
  • ## C, 30 bytes + 1-4 byte
  • `f(m){f(printf("%0*i\n",m,0));}`
  • Has this limitations:
  • - counter stops working at `INT_MAX`, maybe a bit sooner.
  • - Needs a input value to indicate on which line we start. `0` starts on the line with one character.
  • - It uses `0` and not `*` as output character
  • Because of the additional input number, i added the + 1 byte in the title. If you call the program, you need to write `f(0)`, this is why the +4 bytes in the title.
  • ### Fully working program, 36 bytes:
  • `main(m){main(printf("%0*i\n",m,0));}`
  • ## C, 30 bytes + 1-4 byte
  • `f(m){f(printf("%0*i\n",m,0));}`
  • Has this limitations:
  • - counter stops working at `INT_MAX` or at stack overflow.
  • - Needs a input value to indicate on which line we start. `0` starts on the line with one character.
  • - It uses `0` and not `*` as output character
  • Because of the additional input number, i added the + 1 byte in the title. If you call the program, you need to write `f(0)`, this is why the +4 bytes in the title.
  • ### Fully working program, 36 bytes:
  • `main(m){main(printf("%0*i\n",m,0));}`
#1: Initial revision by user avatar H_H‭ · 2023-08-30T14:15:02Z (over 1 year ago)
## C, 30 bytes + 1-4 byte

`f(m){f(printf("%0*i\n",m,0));}`

Has this limitations:

- counter stops working at `INT_MAX`, maybe a bit sooner.
- Needs a input value to indicate on which line we start. `0` starts on the line with one character.
- It uses `0` and not `*` as output character

Because of the additional input number, i added the + 1 byte in the title. If you call the program, you need to write `f(0)`, this is why the +4 bytes in the title.

### Fully working program, 36 bytes: 

`main(m){main(printf("%0*i\n",m,0));}`