Why the FUCK if I remove the last line the "in" char pointer gets the input value?
Actuall, why the FUCK "in" needs to get the input AND NOT the output?
Why the FUCK if I remove the last line the "in" char pointer gets the input value?
Actuall, why the FUCK "in" needs to get the input AND NOT the output?
*Why the FUCK I need to KEEP the last line for the "in" char pointer to get the input value?
Shouldn't it be char[]* for your parameter? Only guessing if you're doing strlen
Why the fuck do you write
>char* = &var[0]
and not
>char* = var
Why?
The parameter doesn't need to be a pointer to a pointer.
Because I am getting the address of the first char in the char array?
but
>&var[0]
is the litteraly the same like
>var
strlen(in) * 4 + 1
Stupid Cniles
No, it's not.
This doesn't answer my question.
>No, it's not.
>No, it's not.
Somebody doesn't know how pointers work.
YOU DIDN'T ANSWER THE QUESTION, YOU FUCKS!
WHY A FUCKING LOG MAKE THE "vi" CHAR POINTER GET THE VALUE FROM THE "output" CHAR POINTER?
bump
just give up lol
No.
Pathetic.
You are.
Can't even answer a simple question.
Oh no no no no
Shouldn't your char arrays be created using malloc?
wait why the fuck you putting success message in the error log?
>char input[strlen(in)];
lmfao
user u have no idea what ur doing, please try using Rust or JS
This is called a single-dimensional array. The arraySize must be an integer constant greater than zero and type can be any valid C data ty