I am running the following C code, where trying to read in buffer which is allocated on caller's stack, but fails with errno 14 (Bad Address).
#include <unistd.h>
#include <fcntl.h>
#include <stdio.h>
#include <errno.h>
#include <stdlib.h>
#include <string.h>
void wrapper(int fd, char **buf)
{
int res = read(fd, *buf, 10);
printf("res: %d, errno: %d\n", res, errno);
printf("Buf: %s\n", *buf);
}
int main()
{
char buffer[10];
memset(buffer, 0, 10);
int fd = open("main.c", O_RDONLY);
wrapper(fd, (char **)&buffer);
return 0;
}
The output is
res: -1, errno: 14
Buf: (null)
I have been searching for explanation why it fails, whereas changing it to
void wrapper(int fd, char *buf)
...
wrapper(fd, (char *)buffer);
works, but without result so far.
why it fails
Arrays are not pointers. buffer
is not a char*
. It's a char[10]
. It will decay to char*
in many contexts, but not in the specific context of taking its address. Consequently, &buffer
is not a char**
, is not compatible with char**
, and should not be cast to char**
. If it is cast to char**
and then dereferenced, the behaviour is undefined.