Write security code-don't compare Structure_structure with memcmp

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Author: gfree.wind@gmail.com
Blog: linuxfocus.blog.chinaunix.net


Take a look at the code below
#include <stdio.h>
#include <stdlib.h>
#include <string.h>


typedef struct PADDING_TYPE {
Short M1;
int m2;
} padding_type_t;


int main ()
{
padding_type_t A = {
. M1 = 0,
. m2 = 0,
};
padding_type_t b;


memset (&b, 0, sizeof (b));


if (0 = = memcmp (&a, &b, sizeof (a)) {
printf ("equal!\n");
}
else {
printf ("No equal!\n");
}


return 0;


}
Let's think about what the result is.
laptop:~/works/test$ gcc-g test.c
laptop:~/works/test$./a.out
No equal!
Why is that so? In fact, experienced development will immediately respond, this is due to the alignment caused.
Yes, because the type of the struct PADDING_TYPE-&GT;M1 is short, and the M2 type is int, according to the principle of natural alignment. Each member of the Padding_type needs to be aligned to 4 bytes. The compiler then inserts 2 padding bytes after the M1, and the padding byte values are random. That is, the value of the padding byte in A is random, while the padding in B is zeroed. So when using Memcmpy to compare the two structures, the return value is unequal.


From this example, we should keep in mind that when comparing structures, do not use byte comparisons, such as memcmp. Unless you're the one to ensure that these aligned padding bytes are zeroed out. Otherwise, you will get unexpected results.