Question

C programming:

What is the difference between the following pieces of code:\[\small\text{char* string = "The quick brown fox jumps over the lazy dog."}\]\[\small\text{char string[] = "The quick brown fox jumps over the lazy dog."}\]

Answer 1

Well, all I know is that the first one is immutable or something., and that the second one is an array of characters with a '\0' in the end, but I'm sure there are deeper differences, both semantically and at the assembly level.

Answer 2

So, there are a lot of differences here. The first, for example, is a pointer while the second is a concrete array. So, one is allocated in the heap and one is allocated in the stack.

Additionally, I'm not entirely sure those assignments are valid, since technically the value for a char* is a memory address, but as I've mentioned before I'm not a regular C programmer.

Answer 3

Actually, those are both valid, both allocated on the stack, and (wait for it).... Both equivalent in C.

They are, for all intents and purposes, exactly the same declaration.

Answer 4

There's a slight possibility that assignments to a char[] might get a warning if you try to assign a char* to it, but I don't think so.

Answer 5

I've done the following:

char* a = "The quick brown fox jumps over the lazy dog.\n";
char b[] = "The quick brown fox jumps over the lazy dog.\n";
a[2] = '\0';
b[2] = '\0';
fprintf(stdout, "%s%s\n", a, b);

And here is the output (note that the first line is a, and the second is b):

The quick brown fox jumps over the lazy dog.
Th;

Answer 6

oops no semicolon on the second one, but you get the difference :-D

Answer 7

I guess there are subtleties that are only apparent once you look under the hood, at the assembly code.

Answer 8

Actually, the subtlety is in C. Both a[2] and b[2] refer to a memory location 2 * sizeof( char ) after it's respective variable. What's goofing you up is that a refers to the pointer itself, it was never dereferenced: *a[2] ( if that syntax is dead nuts, I'm surprised, but you get the idea ).

Answer 9

@rsmith

So *a[2] achieves the intended effect of turning the 3rd character in the sentence into a '\0'? Doesn't work: http://codepad.org/Xc71FImM

Answer 10

For reference here is the original: http://codepad.org/HNPrEHDu

Answer 11

I think it will be *(a + 2) that will do it. Alternatively, *(a + 2 * sizeof(char)), but as I recall the C compiler sees pointer math and does the sizeof multiplication. May be wrong about that though.

Answer 12

I thought a[2] was syntactic sugar for *(a + 2), so it wouldn't work anyway :-P http://codepad.org/VHMfbD7q

*(a + 2 * sizeof(char)) doesn't work either, since sizeof(char) is 1 byte :-P http://codepad.org/YRAbhoSy

Answer 13

I'm surprised it even compiled. *a[2] was just a conceptual description. The upshot is that a[2] would be 2 * sizeof(char) bytes after where the pointer was stored, as opposed to 2 * sizeof(char) bytes after where the pointer pointed to, which would (if the pointer is properly initialized), b[0].