Can anyone explain to me why this isn't working?
#include <stdio.h>
#include <stdlib.h>
char *getline(int lim)
{
char c;
int i;
char *line;
line = malloc(sizeof(char) * lim);
i = 0;
while((c = getchar()) != '\n' && c != EOF && i < lim-1)
{
*line = c;
line++;
i++;
}
*line = '\0';
printf("%s", line);
return line;
}
I'm not worried about the return value right now - just the reason as to why printf("%s", line)
isn't working.
Thanks!
EDIT: fixed to line = malloc(sizeof(char) * lim);
but it is still not working.
Solution: the address of *line
was being incremented throughout the function. When it was passed to printf()
, *line
pointed to '\0' because that's where its adress was incremented to. Using a temprorary pointer that stored the original address allocated by malloc()
to *line
and then passing that pointer into printf()
, allowed for the function to walk up the pointer.
Because you are only allocating enough space for a single character in this line:
And that is getting filled with the
\0
before yourprintf
statement.I'm guessing you want to change this line to:
Or even better:
And then use
tmp
in all of your pointer math, this wayline
will still point to the start of your string.And I know it's early in your development, but you'll want to make sure you
free()
the memory that youmalloc()
.Here is a working version of your function including my suggested changes: