Weird value when I declare a string in C

I want to declare a string that will hold 4 chars 开发者_运维技巧

char st[4];
memcpy(st,"test",4);

but when I print st ... I have "test" and some other symbols printed after it - what is wrong in that ?

Thanks a lot

C strings, like the "test" string literal, are NUL-terminated, meaning the last character is '\0':

{'t', 'e', 's', 't', '\0'}

You would need to use st[5], and copy 5 characters, to have room for (and include) the NUL. As is, you're not including it in the copy. So aftewards, st looks like:

{'t', 'e', 's', 't', X, X, X ... '\0'}

When you print, C keeps reading gibberish values that are coincidentally in memory (X'es above) until it finds a NUL.

The best solution is to eliminate memcpy, and let the compiler figure out the size from your initialization:

char st[] = "test";

sizeof(st) will now be 5.

If you really want to use memcpy(), you have to copy the terminating null-byte as well:

char st[5];
memcpy(st, "test", 5);

remember, in memory "test" looks like that:

74 65 73 74 00
t  e  s  t  \0

that's why you have to copy 5 bytes.

If you don't copy the null byte, functions that work on null terminated strings, such as printf(), will read the memory until they reach some random null-byte...

If for some weird reason want to keep your code as is, do the following:

char st[4];
memcpy(st,"test",4);
int i;
for(i = 0; i < 4; i++)
   printf("%c",st[i]);

Now it will print ok. But read the other answers to see that your programming is not ok.