C - Difference between "char var[]" and "char *var"?

Question

I am expecting that both following vectors have the same representation in RAM:

char a_var[] = "XXX\x00";
char *p_var  = "XXX";

But strange, a call to a library function of type f(char argument[]) crushs the running application if I call it using f(p_var). But using f(a_var) is Ok!

Why?

score 30 · Accepted Answer · answered Dec 10 '09 at 12:14

30

The first creates an array of char containing the string. The contents of the array can be modified. The second creates a character pointer which points to a string literal. String literals cannot be modified.

answered Dec 10 '09 at 12:14

As I know, in ANSI C all can be modified – psihodelia Dec 10 '09 at 12:19
15

You know wrong. The standard forbids modifications of string literals. – Dec 10 '09 at 12:20
4

@psihodelia: Modifying a String literal invokes UB(Undefined Behaviour). – Prasoon Saurav Dec 10 '09 at 12:23
4

I don't have access to the C standard but the C++ standard states (2.13.4.2) "The effect of attempting to modify a string literal is undefined." – Andreas Brinck Dec 10 '09 at 12:30
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@Andreas: The C standard says the same. Read section 6.4.5 :) – Prasoon Saurav Dec 10 '09 at 12:38
The first also performs an implicit copy operation, whereas the second doesn't. This is important on platforms that are memory constrained. – Thomas Matthews Dec 10 '09 at 22:36
1

@Thomas Not true, I'm afraid. The compiler will create the array contents at compile time, – Dec 10 '09 at 22:42
The standard cannot *forbid* things (perhaps only grammar/syntax violations). Saying "invokes undefined behavior" seems more correct. – MestreLion Feb 26 '15 at 11:48

score 1 · Answer 2 · answered Dec 10 '09 at 12:13

1

At a guess, the function f modifies the contents of the string passed to it.

answered Dec 10 '09 at 12:13

Steve Jessop

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yes, it modifies. but why f(p_var) makes crush and f(a_var) not? – psihodelia Dec 10 '09 at 12:14
1

because: char* p_var = "XXX"; Can't be modified. – Pablo Santa Cruz Dec 10 '09 at 12:16

score 1 · Answer 3 · answered Dec 10 '09 at 12:20

Arrays can be treated (generally) as pointers but that doesn't mean that they are always interchangeable. As the other said, your p_var points to a literal, something static that cannot be changed. It can point to something else (e.g. p_var = &a_var[0]) but you can't change the original value that you specified by quotes....

A similar problem is when you are define a variable as an array in one file, and then extern-use it as a pointer.

Regards

score 1 · Answer 4 · answered Oct 25 '11 at 11:58

As others said, char *p_var = "XXX"; creates a pointer to a string literal that can't be changed, so compiler implementations are free to reuse literals, for example:

char *p_var  = "XXX";
char *other  = "XXX";

A compiler could choose to optimize this by storing "XXX" only once in memory and making both pointers point to it, modifying their value could lead to unexpected behavior, so that's why you should not try to modify their contents.

C - Difference between "char var[]" and "char *var"?

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