Why is "Hello World" const char* ?

[Hans]

Hello,

Why all C/C++ guys write:

const char* str = "Hello";

or

const char[] str = "Hello";

Why is this not a more elegant way:

const unsigned char* str = "Hello"; (or [])

??

My Reasson: have you ever seen ASCII-Code: -50? The extended ASCII Set
(>127) would be represented by a negative number if you use 'char'
instead of 'unsigned char'.

Thanks!

Hans

[CBFalconer]

Ron Natalie wrote:

Luke Wu wrote:

In C, a string literal (like "Hello") in value context (like the
right side of an assignment= operator) evaluates to a pointer
value of type (const char*) that points to the first character.

Nope, it evaluates to an array of char, which converts to pointer
to char. The characters are effectively const (it is undefined
behvaior to attempt to change them), but the type of the
expression is not const char*. (except when sizeof or & is
applied to the array value).

It's effectively the same in C++, except C++ defines a formal
array-to-pointer conversion.

However it is advisable to declare such strings as:

const char *foo = "foobar";

so that the C compiler can emit suitable warnings on misuse, such
as trying to alter it with such things as *foo = 'F'; Note that
foo = "barfoo"; is perfectly legal, except you may never find the
"foobar" string again.

--
"If you want to post a followup via groups.google.c om, don't use
the broken "Reply" link at the bottom of the article. Click on
"show options" at the top of the article, then click on the
"Reply" at the bottom of the article headers." - Keith Thompson

[msalters]

Ron Natalie wrote:

Luke Wu wrote:

In C, a string literal (like "Hello") in value context (like the right side of an assignment= operator) evaluates to a pointer value of type (const char*) that points to the first character.
Nope, it evaluates to an array of char, which converts to pointer to
char. The characters are effectively const (it is undefined

behvaior to attempt to change them), but the type of the expression is not
const char*. (except when sizeof or & is applied to the array value).
It's effectively the same in C++, except C++ defines a formal
array-to-pointer conversion.

No, it isn't. In C++ it is definitely an array of N const char, see
2.13.4
and overloading will show that. There is a deprecated limited
conversion
to non-const char*.
Furthermore, even in sizeof( "literal") the type is not const char*,
this will be the true size of the literal object.

( restricted to clc++)

HTH,
Michiel Salters

[Lawrence Kirby]

On Sat, 05 Feb 2005 12:18:22 -0800, john_bode wrote:

Hans wrote:

Hello,

Why all C/C++ guys write:

const char* str = "Hello";

C++ guys don't use char arrays or pointers to char for text data if
they can help it; that's what the std::string datatype is for.

or

const char[] str = "Hello";

Why is this not a more elegant way:

const unsigned char* str = "Hello"; (or [])

Because the type of a string literal is "const char *", not "const
unsigned char*".

As others have said its type is array of char. But the fact that it is
char and not unsigned char is significant.

Another thing that is just as significant is the fact that standard
library functions that take or generate strings have arguments/return
values derived from type char, not unsigned char.
Secondly, the plain, unqualified char type may be signed or unsigned, or
a mixture of both, depending on the implementation. The char type is
not treated exactly the same as other signed and unsigned integral
types.

I'm not sure what you mean by a "mixture of both". For any particular
implementation plain char must behave exclusively like either signed char
or unsigned char, except that it is a different type. Think of it like int
and long. On some implementations they have the same representation but
even when that is true they are different types.

Lawrence

[Ron Natalie]

msalters wrote:

Ron Natalie wrote:

Luke Wu wrote:

In C, a string literal (like "Hello") in value context (like the
right
side of an assignment= operator) evaluates to a pointer value of
type
(const char*) that points to the first character.

Nope, it evaluates to an array of char, which converts to pointer to
char. The characters are effectively const (it is undefined

behvaior

to attempt to change them), but the type of the expression is not
const char*. (except when sizeof or & is applied to the array

value).

It's effectively the same in C++, except C++ defines a formal
array-to-pointer conversion.

No, it isn't. In C++ it is definitely an array of N const char, see
2.13.4

You're right of course. When I wrote the "same in C++" I was refering
to the type being an ARRAY, not a pointer as the previous poster referred to.