memcpy for copying a fixed length buffer into a structure

Question

I have this sample code:

struct
{
    char a[2];
    char b[2];
} buff;

char buffer1[5] = "ABCD";

To copy the buffer1 to structure members, I am doing like this:

char c[3],d[3];
memcpy(&buff,buffer1,4);
sprintf(c,"%2.2s",buff.a);
sprintf(d,"%2.2s",buff.b);
printf("c=%s,d=%s",c,d);

When I print the variables c and d, I am getting the values in c and d variable properly as: c="AB" and c="CD".

Well my question is, even though I am getting the output properly, will memcpy affect anything related to null character termination or have other unexpected memory-related consequences?

Answer 1

The are two issues here:

1) As mentioned in comments you likely forget to include space for the ending '\0' (i.e. NUL in ASCII) terminator character. The %s format specifier for printf function expects that character string is in valid form. However nothing stops you from printing as sequence of characters, like here:

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

struct {
    char a[2];
    char b[2];
} buff;

char buffer1[5] = "ABCD";

int main(void)
{
    memcpy(&buff, buffer1, 4);

    printf("First member: %c%c\n", buff.a[0], buff.a[1]);
    printf("Second member: %c%c\n", buff.b[0], buff.b[1]);

    return 0;
}

2) A more serious issue is that compiler may include arbitrary padding between struct members (as well as after last member), so memcpy may not work as expected. Instead of copying like that (as it may put bytes from array into a unused "wholes"). I would suggest individual copies of each member or maybe using offsetof() macro.

From N1570 6.7.2.1/15 Structure and union specifiers:

There may be unnamed padding within a structure object, but not at its beginning.

and 6.7.2.1/17:

There may be unnamed padding at the end of a structure or union.

Hence, you should split your memcpy into two calls, for instance:

memcpy(&buff.a, buffer1, 2); /* or replace 2 with sizeof buff.a */
memcpy(&buff.b, buffer1+2, 2);