Post History
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...
Answer
#4: Post edited
## 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
- ## 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
- ## 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
## 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));}`