memcpy using pointers throwing access violation

Question

I have a very simple c question regarding memcopy that I have been able to fix but not understand why. The following code crashes with an "access violation" at runtuime:

char *src = "HELLO WORLD!";
char * dest = "hello world!";
memcpy(dest, src, strlen(src)+1);

dest would appear to be big enough and as I understand it memcpy should just blindly copy the bytes so I don't understand the problem.

Changing dest to an array e.g char dest[13]; fixes things, so is it a requirement that dest must be uninitialized memory and creating an array does this for you whereas the pointer declaration does not?

Cheers

Literal strings do not work this way. Create new buffers. – SJuan76 Mar 15 '14 at 16:17 — SJuan76, Mar 15 '14 at 16:17

score 12 · Accepted Answer · answered Mar 15 '14 at 16:19

When you do:

char *dest = "hello world!";

"hello world!" is kept in the read-only parts of the memory and pointer named dest is pointed to it, making any writing operation on this memory illegal. So if you try to use it as a destination in your memcpy; it's an access violation.

You should use:

char dest[] = "hello world!";

user3392484 · Answer 2 · 2014-03-15T17:09:06.513

3

A string literal, "...", when used to initialise a char pointer, has type const char *, i.e. it's read-only.

dest should therefore have type const char *, at which point your code won't compile.

If you turn all your compiler warnings on then you'll get a warning for the "char * dest = ..." line.

edited Mar 15 '14 at 17:09

answered Mar 15 '14 at 16:18

user3392484

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"...." is `char[]` and it is read only because it resides in read only section. And `char* arr = malloc(6); const char* arrptr= arr; memcpy(arrptr,"hello",5); printf("%s",arrptr);` compiles without warnings at my machine – Dabo Mar 15 '14 at 16:40
char foo[] = "..." and char *foo = "..." aren't the same thing. Regarding compiler warnings: gcc 4.8 with default warning options says "passing argument 1 of memcpy discards const qualifier", which is precisely the problem. Either investigate turning on more warnings, or (if possible) get a better compiler. It's also worth compiling with multiple compilers, because they all spot different things. – user3392484 Mar 15 '14 at 16:56
Regarding const-ness of "...": C99 section 6.4.5, paras 5, 6: "The multibyte character sequence is then used to initialize an array ... If the program attempts to modify such an array, the behavior is undefined." – user3392484 Mar 15 '14 at 16:58
Regarding type of string literal read this answer http://stackoverflow.com/a/2245983/2549281 . – Dabo Mar 15 '14 at 17:04
Yes, but it's used in an initialiser, so you need also to read section 6.7.8 of the standard. In that char[] case you're initialising an array (para 14); in the char * case you're initialising a scalar (para 11). – user3392484 Mar 15 '14 at 17:07

score 1 · Answer 3 · answered Mar 15 '14 at 16:25

char dest [] = "hello world!" and char* dest = "hello world!" is not the same thing. char dest[]="hello world!" is an array with 13 elements (the string is null-terminated). char* dest ="hello world!" is a pointer to an constant, null terminated string (also 13 characters).

memcpy using pointers throwing access violation

3 Answers3