String literals non const?

[Marcin Kalicinski]

Why string literals are regarded as char * not as const char *?

(1) void f(char *);
(2) void f(const char *);

f("foo") will call version (1) of function f.

I understand that the exact type of literal "foo" is char[4], which by means
of standard conversion becomes char *. Still, there's something going wrong
here, because this allows modification of "foo", which in my opinion should
be forbidden (because it causes undefined behavior).

Best regards,
Marcin

[Jacques Labuschagne]

Marcin Kalicinski wrote:

Why string literals are regarded as char * not as const char *?

(1) void f(char *);
(2) void f(const char *);

f("foo") will call version (1) of function f.

I understand that the exact type of literal "foo" is char[4], which by means
of standard conversion becomes char *. Still, there's something going wrong
here, because this allows modification of "foo", which in my opinion should
be forbidden (because it causes undefined behavior).

Best regards,
Marcin

It should be const. Section 2.13.4.1 states
``An ordinary string literal has type "array of n const char" and a
static storage duration where n is the size of the string as
defined below, and is initialized with the given characters.''

HTH,
Jacques

[Marcin Kalicinski]

Hi,

So it seems there's something wrong with VC++ .NET compiler. The following
code:

//=============== ===========
#include <iostream>
void main()
{
std::cout << typeid("foo").n ame() << "\n";
}
//=============== ===========

Outputs "char[4]". And the following one:

//=============== ===========
#include <iostream>
int f(char *) { return 1; }
int f(const char *) { return 2; }
void main()
{
std::cout << f("foo") << "\n";
}
//=============== ===========

Outputs 1.

At the moment I do not have access to any other compiler, but I wonder if
this is normal behavior? If the standard says it should be const char [4],
it's a serious problem if this is char[4] instead.

Best regards,
Marcin

Marcin Kalicinski wrote:

Why string literals are regarded as char * not as const char *?

(1) void f(char *);
(2) void f(const char *);

f("foo") will call version (1) of function f.

I understand that the exact type of literal "foo" is char[4], which by means of standard conversion becomes char *. Still, there's something going wrong here, because this allows modification of "foo", which in my opinion should be forbidden (because it causes undefined behavior).

Best regards,
Marcin

It should be const. Section 2.13.4.1 states
``An ordinary string literal has type "array of n const char" and a
static storage duration where n is the size of the string as
defined below, and is initialized with the given characters.''

HTH,
Jacques

[Jacques Labuschagne]

Marcin Kalicinski wrote:

Hi,

So it seems there's something wrong with VC++ .NET compiler. The following
code:

//=============== ===========
#include <iostream>
void main()
{
std::cout << typeid("foo").n ame() << "\n";
}
//=============== ===========

Outputs "char[4]". And the following one:

Interestingly enough, GCC 3.2.3 doesn't show the difference between
char[] and const char[] through typeid.name() -
cout << typeid(char[4]).name() << '\n'
<< typeid(const char[4]).name();
says
A4_c
A4_c

It calls the correct function though.. it matches void f(const char*).

HTH,
Jacques.

[Sumit Rajan]

"Marcin Kalicinski" <ka****@poczta. onet.pl> wrote in message
news:bt******** **@korweta.task .gda.pl...

//=============== ===========
#include <iostream>
int f(char *) { return 1; }
int f(const char *) { return 2; }
void main()
{
std::cout << f("foo") << "\n";
}
//=============== ===========

Outputs 1.

On changing void main() to int main() and compiling using Comeau C++, the
output is 2. On mingw, the output is 2.

Curiously, the output is 1 on Borland 5.5.1.

Regards,
Sumit.

[Magnus]

"Marcin Kalicinski" <ka****@poczta. onet.pl> skrev i melding
news:bt******** **@korweta.task .gda.pl...

Hi,

So it seems there's something wrong with VC++ .NET compiler. The following
code:

//=============== ===========
#include <iostream>
void main()
{
std::cout << typeid("foo").n ame() << "\n";
}
//=============== ===========

Outputs "char[4]". And the following one:

//=============== ===========
#include <iostream>
int f(char *) { return 1; }
int f(const char *) { return 2; }
void main()
{
std::cout << f("foo") << "\n";
}
//=============== ===========

This outputs 2 on Visual Studio .NET 2003 or VC++ 7.1

- Magnus

[Marcin Kalicinski]

Hi,

I tested it on VC .NET (not 2003), compiler version 13.00.9466. Seems that
Microsoft are fixing their bugs :-)

Best regards,
Marcin

U¿ytkownik "Magnus" <[e-mail]> napisa³ w wiadomoœci
news:3f******** @news.broadpark .no...

"Marcin Kalicinski" <ka****@poczta. onet.pl> skrev i melding
news:bt******** **@korweta.task .gda.pl...

Hi,

So it seems there's something wrong with VC++ .NET compiler. The following code:

//=============== ===========
#include <iostream>
void main()
{
std::cout << typeid("foo").n ame() << "\n";
}
//=============== ===========

Outputs "char[4]". And the following one:

//=============== ===========
#include <iostream>
int f(char *) { return 1; }
int f(const char *) { return 2; }
void main()
{
std::cout << f("foo") << "\n";
}
//=============== ===========

This outputs 2 on Visual Studio .NET 2003 or VC++ 7.1

- Magnus

[Ron Natalie]

"Marcin Kalicinski" <ka****@poczta. onet.pl> wrote in message news:bt******** **@korweta.task .gda.pl...

Why string literals are regarded as char * not as const char *?

(1) void f(char *);
(2) void f(const char *);

f("foo") will call version (1) of function f.

Your compiler is broken. There are two "exact match"
conversion sequences from string literal to char* and const char*.
The overloads above are ambiguous.

There is a deprecated conversion from string literal to char* which exists to
codify 25 years of sloppy C programming.

[David White]

"Jacques Labuschagne" <ja*****@clawsh rimp.com> wrote in message
news:qt******** ************@ne ws02.tsnz.net.. .

Marcin Kalicinski wrote:

Why string literals are regarded as char * not as const char *?

(1) void f(char *);
(2) void f(const char *);

f("foo") will call version (1) of function f.

I understand that the exact type of literal "foo" is char[4], which by means of standard conversion becomes char *. Still, there's something going wrong here, because this allows modification of "foo", which in my opinion should be forbidden (because it causes undefined behavior).
The conversion to char * is allowed for compatibility with lots of old code
that uses char * as though it were const char *.
Best regards,
Marcin

It should be const. Section 2.13.4.1 states
``An ordinary string literal has type "array of n const char" and a
static storage duration where n is the size of the string as
defined below, and is initialized with the given characters.''

There's also 4.2.2:
"A string literal (2.13.4) that is not a wide string literal can be
converted to an rvalue of type "pointer to
char"; a wide string literal can be converted to an rvalue of type "pointer
to wchar_t". In either case,
the result is a pointer to the first element of the array. This conversion
is considered only when there is an
explicit appropriate pointer target type, and not when there is a general
need to convert from an lvalue to an rvalue. [Note: this conversion is
deprecated. See Annex D. ] For the purpose of ranking in overload resolution
(13.3.3.1.1), this conversion is considered an array to pointer conversion
followed by a qualification conversion (4.4). [Example: "abc" is converted
to "pointer to const char" as an array to pointer conversion, and then to
"pointer to char" as a qualification conversion. ]

DW

[Chris Mantoulidis]

"Marcin Kalicinski" <ka****@poczta. onet.pl> wrote in message news:<bt******* ***@korweta.tas k.gda.pl>...

Hi,

So it seems there's something wrong with VC++ .NET compiler. The following
code:

M$ products, BAH!!!!

My GCC 3.3.2 will work fine with that and call the const char * version


//=============== ===========
#include <iostream>
void main()
{
std::cout << typeid("foo").n ame() << "\n";
}
//=============== ===========

Outputs "char[4]". And the following one:

//=============== ===========
#include <iostream>
int f(char *) { return 1; }
int f(const char *) { return 2; }
void main()
{
std::cout << f("foo") << "\n";
}
//=============== ===========

Outputs 1.

At the moment I do not have access to any other compiler, but I wonder if
this is normal behavior? If the standard says it should be const char [4],
it's a serious problem if this is char[4] instead.

Best regards,
Marcin

Marcin Kalicinski wrote:

Why string literals are regarded as char * not as const char *?

(1) void f(char *);
(2) void f(const char *);

f("foo") will call version (1) of function f.

I understand that the exact type of literal "foo" is char[4], which by means of standard conversion becomes char *. Still, there's something going wrong here, because this allows modification of "foo", which in my opinion should be forbidden (because it causes undefined behavior).

Best regards,
Marcin

It should be const. Section 2.13.4.1 states
``An ordinary string literal has type "array of n const char" and a
static storage duration where n is the size of the string as
defined below, and is initialized with the given characters.''

HTH,
Jacques