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const char* = new char[6]

Hi Everyone

I have

const char *p = "Hello";

So, here memory is not allocated by C++ compiler for p and hence I
cannot access p[0] to modify the contents to "Kello"
p[0] = 'K'; // error at runtime

So I did

const char *p = new char[6];

But then how do I initialize it to "Hello"? My requirement is - I want
a const char* initialised and later want to modify the contents.

I know a way as written below

const char p[] = "hello";
const_cast<char &>(p[0]) = 'K'; //OK

But how to acheive this with pointers?

What I know is - C compiler allocates memory when I do
const char* = "hello";

But C++ compiler does not do that. Any help is welcome.


Oct 3 '06
42 32204
S S posted:
A const* p = new A const[3](5);

What compiler are you using? G++ gives:

ISO C++ forbids initialization in array new

Frederick Gotham
Oct 5 '06 #41
S S posted:
Even if you do not put brackets () and write
int *p = new int[3];
then also it does the default initialisation for all 3 members of
array, any significance of brackets?

The Standard provides no such guarantee (well I'm 99% that it doesn't) --
if you want each element to be default-initialised, you'll have to use the
empty parentheses.

Testing it will do you no good, as I know of at least one implementation
that default-initialises new'ed memory regardless of whether you provide
the empty parentheses.

Anyhow, even if the empty parentheses were redundant, I would still put
them in, just as how I write:

int main()
static MyPOD arr[5] = {};

instead of:

int main()
static MyPOD arr[5];

Indeed, the chain brackets are redundant because static data always gets
default initialised... but nonetheless, they express clear intent.


Frederick Gotham
Oct 5 '06 #42

S S wrote:
Frederick Gotham wrote:
int *p = new int[3]();

Even if you do not put brackets () and write
int *p = new int[3];
then also it does the default initialisation for all 3 members of
array, any significance of brackets?
You are mistaken. No initialization occurs if the parens are not
there. Not even default initialization.

Oct 5 '06 #43

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