C char array as pointers

Question

I want to understand:

• why it happens that sometimes a char[1] in C is used as char* (why doing this?) and
• how the internals works (what's going on)

Giving following sample program:

#include <stdio.h>
#include <string.h>

struct test_struct {
    char *a;
    char b[1];
} __attribute__((packed)); ;

int main() {

    char *testp;
    struct test_struct test_s;

    testp = NULL;
    memset(&test_s, 0, sizeof(struct test_struct));

    printf("sizeof(test_struct) is: %lx\n", sizeof(struct test_struct));

    printf("testp at: %p\n", &testp);
    printf("testp is: %p\n", testp);

    printf("test_s.a at: %p\n", &test_s.a);
    printf("test_s.a is: %p\n", test_s.a);

    printf("test_s.b at: %p\n", &test_s.b);
    printf("test_s.b is: %p\n", test_s.b);

    printf("sizeof(test_s.b): %lx \n", sizeof(test_s.b));

    printf("real sizeof(test_s.b): %lx \n", ((void *)(&test_s.b) - (void *)(&test_s.a)) );

    return 0;
}

I get the following output (OS X, 64bit):

sizeof(test_struct) is: 9
testp at: 0x7fff62211a98
testp is: 0x0
test_s.a at: 0x7fff62211a88
test_s.a is: 0x0
test_s.b at: 0x7fff62211a90
test_s.b is: 0x7fff62211a90
sizeof(test_s.b): 1 
real sizeof(test_s.b): 8 

Looking at the memory addresses, one can see that even the struct is 9 bytes large, 16 bytes were allocated which seems to be caused by char b[1]. But I'm not sure if those extra bytes were allocated due to optimization/mem alignment reasons, or if this has to do with C's internal treatment of char arrays.

A real world example can be seen in <fts.h>:

`man 3 fts` shows the struct member `fts_name` as:

            char *fts_name;                 /* file name */

while /usr/include/fts.h defines the member as:

            char fts_name[1];               /* file name */

In the end, fts_name can really be used as a pointer to a C-string. For example, printing to stdout with printf("%s", ent->fts_name) works.

So if a char[1] is really one byte large, it couldn't be used as a memory pointer on my 64bit machine. On the other hand, treating this as a full blown char * doesn't work either, as can be seen with the test_s.b is output above, which should show a NULL pointer then...

Answer 1

Here is an answer that describes the char[1] trick. Basically, the idea is to allocate more memory when malloc()ing the struct, to already have some storage for your string without additional allocation. You can sometimes even see char something[0] used for the same purpose, which makes even less intuitive sense.

On the other hand, treating this as a full blown char * doesn't work either, as can be seen with the test_s.b is output above, which should show a NULL pointer then...

If something is an array, both its name and &name just give the pointer to the start of the array in C. This works regardless of whether it's a member in a struct, or a free standing variable.

printf("real sizeof(test_s.b): %lx \n", ((void *)(&test_s.b) - (void *)(&test_s.a)) );

This line gives the size of space allocated for a, not b in this struct. Put something after b and used this to subtract. With the packed attribute (which means you disallow the compiler to mess with alignment, etc.), you should get 1.

#include <stdio.h>
#include <string.h>

struct test_struct {
    char *a;
    char b[1];
    char c;
} __attribute__((packed));

int main() {
  struct test_struct s;
  printf("%lx\n", ((void*)&s.c) - ((void*)&s.b));
  return 0;
}

I get 1.