Re: assigning const char* to char*
- From: Keith Thompson <kst-u@xxxxxxx>
- Date: Fri, 01 Jun 2007 13:20:57 -0700
Peithon <Peithon@xxxxxxxxxxxxxx> writes:
This is a very simple question but I couldn't find it in your FAQ.
I'm using VC++ and compiling a C program, using the /TC flag.
I've got a function for comparing two strings
int strspcmp(const char * s1, const char * s2)
{
char* pS1 = s1;
char* pS2 = s2;
...
...
}
As you can see, I assign the two const ptrs to some non-const ptrs to
do some ptr arithmetic,
like incrementing the ptrs to compare the strings.
But I'm getting warning C4090 complaining about the assignment.
I thought I was doing the standard thing for ptr arithmetic. What am I
doing wrong?
How do I get rid of the warning?
This has already been ably answered, but perhaps a simpler example
might make the point clearer. I think your point of confusion is that
"const char*" means "pointer to const char", not "const pointer to
char", but I'll explore other issues as well.
int main(void)
{
const char *ptr = "hello";
char *copy_of_ptr = ptr;
*ptr = 'H';
return 0;
}
The declaration "const char *ptr" doesn't make ptr const; it makes
what it points to const. Here I've initialized ptr to point to a
string literal, i.e., to an array of 6 characters 'h', 'e', 'l', 'l',
'o', '\0'. This array could be stored in read-only memory; attempting
to modify it could cause the program crash (it's undefined behavior).
The declaration of copy_of_ptr doesn't use the "const" keyword, which
means I'm allowed to modify either the pointer object itself *or* what
it points to. By "allowed to", I mean that the compiler won't
complain if I try to do it; it will go ahead and generate code to do
what I asked. It's up to me to make sure that I actually *can* modify
whatever I'm trying to modify. If I lie to the compiler, it may or
may not complain.
But in this case, I *tried* to lie to the compiler and it caught me at
it. I attempted to initialize a char* to the value of ptr. If I
could do this, it would allow me to (attempt to) modify whatever
copy_of_ptr points to. And that's exactly why the language forbids
assigning the value of ptr to copy_of_ptr (it's a constraint
violation, requiring a diagnostic). My error is caught during
compilation, when I attempt to copy the pointer, which is much better
than not catching it until run time, when I actually try to modify the
string.
If the language, and therefore the compiler, allowed me to copy the
value of ptr to copy_of_ptr, then it would also allow me to attempt to
modify the string that it points to (remember that it might be in
read-only memory), all without telling me that I've done something
dangerous. Fortunately, the rules of the language are such that I
can't get away with this particular error quite so easily. (It is
overly lax about string literals, but that's another story.)
If you want a pointer to const char, declare it as:
const char *ptr;
If you want a const pointer to char (you can't modify the pointer, but
you can modify what it points to):
char *const ptr;
And if you want both the pointer *and* what it points to be read-only:
const char *const p;
The placement of the "const" keyword is tricky. If you can find a
program called "cdecl", it can help you with this kind of thing; I
used it to confirm these declarations.
% cdecl
Type `help' or `?' for help
cdecl> declare p as const pointer to char
char * const p
cdecl> declare p as const pointer to const char
const char * const p
cdecl> explain const char *p
declare p as pointer to const char
cdecl>
One thing to watch out for: cdecl treats "ptr" as a synonym for
"pointer", so if you tell it "declare ptr as pointer to const char" it
will report a syntax error.
--
Keith Thompson (The_Other_Keith) kst-u@xxxxxxx <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
.
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