C memory overlap?

Question

I am trying to copy the first 16 bytes of a 32 byte string to dest.

unsigned char src[32] = "HELLO-HELLO-HELLO-HELLO-HELLO-12";
unsigned char dest[16];

memcpy(dest, src, 16); // COPY

printf("%s\n", src);
printf("%lu\n", strlen(src));

printf("%s\n", dest);
printf("%lu\n", strlen(dest));

The output is the following

HELLO-HELLO-HELLO-HELLO-HELLO-12
32
HELLO-HELLO-HELLHELLO-HELLO-HELLO-HELLO-HELLO-12
48

I was expecting to receive HELLO-HELLO-HELL in dest only. The first 16 bytes of dest actually contain the expected result.

Why does dest more than it can actually hold? Why does it have a length of 16+32=48? Is there a way to copy only the first 16 bytes of src to dest?

If you want to copy a 16 character string, you need a 17 character buffer, and ensure that the last character is a `\0`. What's happening here is a classic printing from an unterminated string yielding garbage (which in this case is the remaining characters from the src string). — Anya Shenanigans, Dec 10 '14 at 21:00
Take a look at the [string tag-wiki](https://stackoverflow.com/tags/string/info). — Deduplicator, Dec 10 '14 at 21:01
Use strcpy or strncpy instead of memcpy and read up about how C strings which are not really strings like those in other languages but null terminated array's of char. — hookenz, Dec 10 '14 at 21:05

score 7 · Accepted Answer · answered Dec 10 '14 at 21:01

7

The 16 bytes allocated for dest need to include a byte for the trailing NULL ('\0') -- since you wrote 16 bytes, you have a non-null terminated string.

Since the computer you are on has the stack organized in a particular way, you are proceeding past the end of dest and then printing src.

So, to copy 16 bytes, allocate 17 to leave room for the trailing null, and initialize it.

unsigned char dest[17]={0};

Alternative, after copying it, null terminate it:

memcpy(dest, src, 16); // COPY
dest[16]='\0';

answered Dec 10 '14 at 21:01

JohnH

2,713
12
21

1

Spurious initializations are a good way to kill your performance. It's not too bad in this case, as you only do 17 times the initialization work needed, as the buffer isn't too big. – Deduplicator Dec 10 '14 at 21:03
18 seconds! You win. – KirkSpaziani Dec 10 '14 at 21:03
1

*Spurious* . . . I do not think it means what you think it means. – Mike Sherrill 'Cat Recall' Dec 10 '14 at 21:16

score 3 · Answer 2 · answered Dec 10 '14 at 21:02

You're not accounting for the fact that strings are null terminated. There is a '\0'character at the end of the "HELLO..." string constant. Take a look at

http://www.tutorialspoint.com/c_standard_library/c_function_strncpy.htm

That should point you in the right direction. memcpy doesn't provide help for C Strings - just raw memory.

C memory overlap?

2 Answers2