C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  Simple Question--char *          C++Talk.NET Forum Index -> C++ Language (Moderated) View previous topic :: View next topic   Author Message dave Guest Posted: Tue Feb 07, 2006 1:00 pm    Post subject: Simple Question--char * I think I understand that char * ptr = "string" declares a pointer initialized to a possibly modifiable constant string. So, I thought I knew what I would see when I ran the following: char * char_ptr = "stringthingy"; int integer = 1234; int * int_ptr = &integer; ... cout << "Size of char_ptr is " << sizeof(char_ptr) << "\n\n"; cout << "*char_ptr = " << *char_ptr << "\n"; cout << "char_ptr = " << char_ptr << "\n"; cout << "Size of int_ptr is " << sizeof(int_ptr) << "\n\n"; cout << "*int_ptr = " << *int_ptr << "\n"; cout << "int_ptr = " << int_ptr << "\n"; But, I was surprised to see this result: Size of char_ptr is 4 *char_ptr = s char_ptr = stringthingy Size of int_ptr is 4 *int_ptr = 1234 int_ptr = 0x804e03c Why is the non-deferenced char_ptr not an address in memory?? The non-dereferenced int_ptr result is. Is "<<" overloaded to dereference a char ptr? If so, why not for an int ptr? If it is overloaded, why does "<< char_ptr" produce a different result from "<< *char_ptr"? I am a bit confused. TIA, ~Dave~ [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Kurt Stege Guest Posted: Tue Feb 07, 2006 3:00 pm    Post subject: Re: Simple Question--char * dave wrote: Quote: I think I understand ... I think you understand quite much about pointers and C++. Examples and details to questions snipped... Quote: Is "<<" overloaded to dereference a char ptr? Yes. In a way that the pointer is interpreted as an old fashioned pointer to a zero terminated string. The reason is to allow things like std::cout << "Hello World!\n"; In this context you expect to see the string, and not the address of the string as output. Quote: If so, why not for an int ptr? The reason for char ptr does not work for int ptr. Quote: If it is overloaded, why does "<< char_ptr" produce a different result from "<< *char_ptr"? Because << is overloaded, not *. *char_ptr is of type char. And << called with a char outputs exactly that one character, as expected by std::cout << '\n'; It would be possible to overload << to take the address of the given char and output the following characters in memory as well, but this is far from useful and nobody has done it. Quote: I am a bit confused. That is good. A sign for real understanding. Best regards, Kurt. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Valentin Samko Guest Posted: Wed Feb 08, 2006 11:00 am    Post subject: Re: Simple Question--char * dave wrote: Quote: char * ptr = "string" declares a pointer initialized to a possibly modifiable constant string. It's not modifiable. ptr is modifiable, the string it points to - no. Quote: So, I thought I knew what I would see when I ran the following: char * char_ptr = "stringthingy"; int integer = 1234; int * int_ptr = &integer; cout << "*int_ptr = " << *int_ptr << "\n"; And here you get an undefined behaviour, as you are dereferencing an int ptr which points to a storage area which was not initialised as an integer. Quote: Why is the non-deferenced char_ptr not an address in memory?? The non-dereferenced int_ptr result is. Is "<<" overloaded to dereference a char ptr? Yes Quote: If so, why not for an int ptr? How would you expect it to work for int ptr? Quote: If it is overloaded, why does "<< char_ptr" produce a different result from "<< *char_ptr"? Because in addition to be consistent (char* and char are different types, first points to possibly more than one character, and the second one is always one character), this is required by the standard, see 27.6.2.5.4/4 -- Valentin Samko - http://www.valentinsamko.com [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Alberto Ganesh Barbati Guest Posted: Wed Feb 08, 2006 2:00 pm    Post subject: Re: Simple Question--char * dave ha scritto: Quote: I think I understand that char * ptr = "string" declares a pointer initialized to a possibly modifiable constant string. Your understanding is not entirely correct. 2.13.4/2 says that "The effect of attempting to modify a string literal is undefined." So, if you modify the string, you do it at your own risk: the application may run without errors or ignore the modifications or crash or even format your hard-disk. In fact the type of a string literal is "array of n const char" (notice the const). The quirk is that a string literal can be used to initialize a char*. This is an exception whose use is deprecated and that is kept mainly for backward compatibility with C and older C++ programs. Quote: Why is the non-deferenced char_ptr not an address in memory?? The non-dereferenced int_ptr result is. Is "<<" overloaded to dereference a char ptr? If so, why not for an int ptr? If it is overloaded, why does "<< char_ptr" produce a different result from "<< *char_ptr"? Don't confuse "to be an address in memory" with "to be output on cout as something that resembles an address in memory"! operator<< simply provides a textual representation of the right operand, but what the "textual representation" is can differ according to the type of the argument. As you already realized, operator<< has several overloads to provide textual representation of integers, floats, etc. and among them there are overloads for const char* and for const void*. A const char* is assumed to be pointing to a null-terminated character string, so its textual representation is the characters in the string itself. Compare with the typical statement std::cout << "hello, world\n"; That explains why your char* is output as a string and not as an address. With const void* you can't assume that it points to anything meaningful, so the operator<< behaves as the printf "%p" format specifier which outputs an implementation-defined representation of the pointer (typically, its address in numeric form). For pointer types different from const char* (including signed and unsigned variations) the compiler always choses the overload for const void* and that's why your int* is output as an address. HTH, Ganesh [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top kanze Guest Posted: Wed Feb 08, 2006 3:00 pm    Post subject: Re: Simple Question--char * dave wrote: Quote: I think I understand that char * ptr = "string" declares a pointer initialized to a possibly modifiable constant string. Not quite. The string is not "possibly modifiable"; it's not modifiable at all. The lack of const in the type is a lie, supported for reasons of backward compatibility only; you really should write this: char const* ptr = "string" ; Quote: So, I thought I knew what I would see when I ran the following: char * char_ptr = "stringthingy"; int integer = 1234; int * int_ptr = &integer; ... cout << "Size of char_ptr is " << sizeof(char_ptr) << "\n\n"; cout << "*char_ptr = " << *char_ptr << "\n"; cout << "char_ptr = " << char_ptr << "\n"; cout << "Size of int_ptr is " << sizeof(int_ptr) << "\n\n"; cout << "*int_ptr = " << *int_ptr << "\n"; cout << "int_ptr = " << int_ptr << "\n"; But, I was surprised to see this result: Size of char_ptr is 4 *char_ptr = s char_ptr = stringthingy Size of int_ptr is 4 *int_ptr = 1234 int_ptr = 0x804e03c Why is the non-deferenced char_ptr not an address in memory?? The non-dereferenced int_ptr result is. Is "<<" overloaded to dereference a char ptr? Exactly. (For char const*, actually, but the conversion to char const* is better than the convertion to void const*.) Quote: If so, why not for an int ptr? Because the overload for char const* was necessary to support string literals. You cannot define a function to take a char const[] (the actual type of a string literal) as a parameter; the closest you can come is a function which accepts char const*. Quote: If it is overloaded, why does "<< char_ptr" produce a different result from "<< *char_ptr"? Because the overload of << for char (the result type of *char_tpr) outputs a single char -- which is all it can do, if we also allow "<< '\n'" and such things. The overload of char const*, on the other hand, interprets the address as the start of a '\0' terminated string. -- James Kanze GABI Software Conseils en informatique orientée objet/ Beratung in objektorientierter Datenverarbeitung 9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34 [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Francis Glassborow Guest Posted: Wed Feb 08, 2006 5:00 pm    Post subject: Re: Simple Question--char * In article <gO9Gf.83683$eD5.1412447 (AT) twister2 (DOT) libero.it>, Alberto Ganesh Barbati <AlbertoBarbati (AT) libero (DOT) it> writes Quote: For pointer types different from const char* (including signed and unsigned variations) the compiler always choses the overload for const void* and that's why your int* is output as an address. I suspect not. What if I provide an overload for operator << that takes a mytype* as an argument: ostream & operator << (ostream &, mytype * const &); -- Francis Glassborow ACCU Author of 'You Can Do It!' see http://www.spellen.org/youcandoit For project ideas and contributions: http://www.spellen.org/youcandoit/projects [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Kurt Stege Guest Posted: Wed Feb 08, 2006 8:10 pm    Post subject: Re: Simple Question--char * Valentin Samko wrote: Quote: dave wrote: char * ptr = "string" declares a pointer initialized to a possibly modifiable constant string. It's not modifiable. ptr is modifiable, the string it points to - no. Yes. Here you are more or less right. It is allowed to write the code *ptr='x'; but when executing that line it results in undefined behaviour. I suppose, dave noticed this curious behaviour of C++ and thus called id "possibly modifiable", as "you can but you can't". Quote: So, I thought I knew what I would see when I ran the following: char * char_ptr = "stringthingy"; int integer = 1234; int * int_ptr = &integer; cout << "*int_ptr = " << *int_ptr << "\n"; And here you get an undefined behaviour, as you are dereferencing an int ptr which points to a storage area which was not initialised as an integer. You have noticed the line int_ptr = &integer? Best regards, Kurt. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Valentin Samko Guest Posted: Thu Feb 09, 2006 10:06 am    Post subject: Re: Simple Question--char * Kurt Stege wrote: Quote: char * ptr = "string" Yes. Here you are more or less right. It is allowed to write the code *ptr='x'; but when executing that line it results in undefined behaviour. I wouldn't say that something is "doable" in a language if this results in an undefined behaviour. Otherwise one could say that dereferencing a 0 pointer is allowed in C++. Quote: And here you get an undefined behaviour, as you are dereferencing an int ptr which points to a storage area which was not initialised as an integer. You have noticed the line int_ptr = &integer? No I haven't There is no undefined behaviour there. -- Valentin Samko - http://www.valentinsamko.com [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Alberto Ganesh Barbati Guest Posted: Thu Feb 09, 2006 11:06 am    Post subject: Re: Simple Question--char * Francis Glassborow ha scritto: Quote: In article <gO9Gf.83683$eD5.1412447 (AT) twister2 (DOT) libero.it>, Alberto Ganesh Barbati <AlbertoBarbati (AT) libero (DOT) it> writes For pointer types different from const char* (including signed and unsigned variations) the compiler always choses the overload for const void* and that's why your int* is output as an address. I suspect not. What if I provide an overload for operator << that takes a mytype* as an argument: ostream & operator << (ostream &, mytype * const &); Sure, I should have said "always choses the overload for const void* unless you provide additional overloads". Thanks for pointing that out. Ganesh [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. 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