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String literals non const?

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
Jul 22 '05
10 3636
On Fri, 2 Jan 2004 08:55:33 -0500, "Ron Natalie" <ro*@sensor.com >
wrote:

"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.


No they aren't. string literal -> char* is equivalent to an
array-to-pointer conversion followed by a qualification conversion,
whereas -> const char* is just an array-to-pointer conversion. (2)
will be chosen unambiguously.

Tom

C++ FAQ: http://www.parashift.com/c++-faq-lite/
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
Jul 22 '05 #11

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