C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  How to cast (char *) to (char *)& for a "call by reference" Goto page 1, 2  Next          C++Talk.NET Forum Index -> C++ Language (Moderated) View previous topic :: View next topic   Author Message Jerry Guest Posted: Fri Oct 28, 2005 9:40 am    Post subject: How to cast (char *) to (char *)& for a "call by reference" Here is the program: void foo(char *& Path) { // Do something with szPath... } int main(int argc, char* argv[]) { char PathA[] = "MyPath"; foo(PathA); // This call has compiling error return 0; } I got following error: - error C2664: 'foo' : cannot convert parameter 1 from 'char [7]' to 'char *& ' Can somebody tell me how should I cast the PathA to get it working? Thanks! -J [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Stefan Näwe Guest Posted: Fri Oct 28, 2005 6:27 pm    Post subject: Re: How to cast (char *) to (char *)& for a "call by refere Jerry wrote: Quote: Here is the program: void foo(char *& Path) { // Do something with szPath... } int main(int argc, char* argv[]) { char PathA[] = "MyPath"; char* PathA = 0; // Initialize PathA if needed (depends on what foo() will do) // for example: PathA = strdup("MyPath"); // but don't forget to free(PathA) Quote: foo(PathA); // This call has compiling error return 0; } I got following error: - error C2664: 'foo' : cannot convert parameter 1 from 'char [7]' to 'char *& ' Can somebody tell me how should I cast the PathA to get it working? Don't cast. Think about using a std::string. Stefan -- Stefan Naewe naewe.s_AT_atlas_DOT_de [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Sebastian Redl Guest Posted: Sat Oct 29, 2005 3:47 am    Post subject: Re: How to cast (char *) to (char *)& for a "call by referen [Unrecognized group removed from header -- mod] Jerry wrote: Quote: Here is the program: void foo(char *& Path) { // Do something with szPath... } int main(int argc, char* argv[]) { char PathA[] = "MyPath"; foo(PathA); // This call has compiling error return 0; } foo expects a non-const reference as the argument. Non-const references can only bind to lvalues. However, the result of implicitely converting PathA to (char*) is an rvalue (it has no memory address). To make this call succeed, you need to introduce an extra char* variable: int main(int argc, char *argv[]) { char PathA[] = "MyPath"; char *pPathA = PathA; foo(pPathA); return 0; } Note that this is not necessarily what you want: foo can now change what pPathA points to, but it could change the actual contents of PathA even without the reference. -- Sebastian Redl [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Joe Guest Posted: Sat Oct 29, 2005 1:02 pm    Post subject: Re: How to cast (char *) to (char *)& for a "call by referen Quote: Can somebody tell me how should I cast the PathA to get it working? Thanks! -J The short answer is you can't. You have foo() declared to take a reference to a pointer and no matter how you shake it, pathA[] isn't a pointer. It is an array. If you think about it, the only reason to pass a reference to a pointer would be because you intended to modify that pointer in foo(). What would it mean to modify the beginning of an array? Maybe it would be better to show you the memory layout implied by pointers and arrays (Note: compilers can optimize all sorts of things, so this may not be what you end up with, however it is the implied layout). With an array you get something like: pathA: <ch1><ch2>...<chn> With a pointer you have: pPathA: <address> <address>: <ch1><ch2>...<chn> In the first case, the token pathA refers to the address of the first element of the array, in the pointer case, the token refers to a piece of memory containing the address of the first element. The compiler allows pathA to be treated as a pointer because of C compatibility and the lack of any sort of string type in the base language. That is, "Text" is an array (unless it is used for initialization) and it would be inconvenient to always have to declare a pointer to every text literal you might want. So, if you really need foo() to take a reference to a pointer, then you need to change pathA to: char const * pathA = "Text"; then you can call foo(pathA). Personally, I would consider rewriting foo() to accept a char const * instead of the reference. joe [ 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: Sat Oct 29, 2005 1:02 pm    Post subject: Re: How to cast (char *) to (char *)& for a "call by referen Jerry wrote: Quote: Here is the program: void foo(char *& Path) { // Do something with szPath... Presumably, you mean Path. And presumably, that something means modifying the pointer (and not just what it points to), something along the lines of: Path = new char[ n ] ; Quote: } int main(int argc, char* argv[]) { char PathA[] = "MyPath"; foo(PathA); // This call has compiling error Which is to be expected: where is the pointer which is to be modified. Quote: return 0; } I got following error: - error C2664: 'foo' : cannot convert parameter 1 from 'char [7]' to 'char *& ' Can somebody tell me how should I cast the PathA to get it working? Do you want to? The question above stands: where is the pointer which is to be modified? The normal way of handling this would be something like: char PathA[] = "MyPath" ; char* PathAPtr = PathA ; foo( PathAPtr ) ; Realizing, of course, that after the call to foo, PathAPtr may not be equal to PathA. -- 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 xlxieliang@hotmail.com Guest Posted: Sat Oct 29, 2005 1:03 pm    Post subject: Re: How to cast (char *) to (char *)& for a "call by referen void foo(char *& Path) { // Do something with szPath... } int main(int argc, char* argv[]) { char PathA[] = "MyPath"; foo((char*&)PathA); // This call has compiling error return 0; } foo(((char*&)PathA) instead of foo(PathA); [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Thomas Maeder Guest Posted: Sat Oct 29, 2005 1:07 pm    Post subject: Re: How to cast (char *) to (char *)& for a "call by referen "Jerry" <rrrshop (AT) hotmail (DOT) com> writes: Quote: void foo(char *& Path) { // Do something with szPath... That's hard; there is no szPath declared here. Quote: } int main(int argc, char* argv[]) { char PathA[] = "MyPath"; foo(PathA); // This call has compiling error return 0; } I got following error: - error C2664: 'foo' : cannot convert parameter 1 from 'char [7]' to 'char *& ' Can somebody tell me how should I cast the PathA to get it working? Hardly. PathA is not a pointer. What problem are you trying to solve? [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Neelesh Guest Posted: Sat Oct 29, 2005 1:08 pm    Post subject: Re: How to cast (char *) to (char *)& for a "call by referen Jerry wrote: Quote: Here is the program: void foo(char *& Path) { // Do something with szPath... } int main(int argc, char* argv[]) { char PathA[] = "MyPath"; foo(PathA); // This call has compiling error when PathA is passed to foo, char[] is implicitly converted to char* by generation of a temporary. A temporary can be passed as a const& but not as a non-const& . Hence you need to change the prototype as void foo(char * const & Path)// reference to a constant pointer to a character. Hope this helps. Neelesh [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Allan W Guest Posted: Sat Oct 29, 2005 1:08 pm    Post subject: Re: How to cast (char *) to (char *)& for a "call by referen Jerry wrote: Quote: Here is the program: void foo(char *& Path) { // Do something with szPath... } Hmm, you don't say what szPath is... Quote: int main(int argc, char* argv[]) { char PathA[] = "MyPath"; foo(PathA); // This call has compiling error return 0; } I got following error: - error C2664: 'foo' : cannot convert parameter 1 from 'char [7]' to 'char *& ' Foo's parameter is a reference to a pointer. You're passing in a pointer, but not in a variable that can be referenced. It's the same problem as this: void bar(int &x) { x = 3; } int baz() { bar(3+3); } // Error Here, bar accepts a reference to an integer, so that it can modify the integer. baz passes in an integer expression -- but it isn't an integer variable, so it can't be modified. Your program does the same thing. The parameter to foo needs to be a variable of type char*, so that foo() can modify it. But you're passing in an expression of type char*. There's nothing for foo to modify. Quote: Can somebody tell me how should I cast the PathA to get it working? You can't do it with a cast. You need a variable of type char*. int main(int argc, char* argv[]) { char PathA[] = "MyPath"; char*x = PathA; foo(x); return 0; } // ** OR ** int main(int argc, char* argv[]) { char *PathA = "MyPath"; foo(PathA); return 0; } Hope this explains it. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Carlos Moreno Guest Posted: Sat Oct 29, 2005 1:09 pm    Post subject: Re: How to cast (char *) to (char *)& for a "call by referen Jerry wrote: Quote: Here is the program: void foo(char *& Path) { // Do something with szPath... } int main(int argc, char* argv[]) { char PathA[] = "MyPath"; foo(PathA); // This call has compiling error return 0; } I got following error: - error C2664: 'foo' : cannot convert parameter 1 from 'char [7]' to 'char *& ' Can somebody tell me how should I cast the PathA to get it working? Well, what you're doing definitely seems wrong. Don't know if the following is even an option for you, but it works: void foo (char * const & path) As I said, maybe changing the function's signature is not an option for you, but as a reference to const, implicit conversion is allowed. In your original code, because it is a reference to a 'char *', you require an actual char * to refer to, which the array is not. HTH, Carlos -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top John Carson Guest Posted: Sat Oct 29, 2005 1:09 pm    Post subject: Re: How to cast (char *) to (char *)& for a "call by referen "Jerry" <rrrshop (AT) hotmail (DOT) com> wrote Quote: Here is the program: void foo(char *& Path) { // Do something with szPath... } Don't cast. Change it to void foo(char * const & Path) { // Do something with szPath... } Quote: int main(int argc, char* argv[]) { char PathA[] = "MyPath"; foo(PathA); // This call has compiling error return 0; } I got following error: - error C2664: 'foo' : cannot convert parameter 1 from 'char [7]' to 'char *& ' Can somebody tell me how should I cast the PathA to get it working? The problem is that when you make the function parameter a reference to a pointer, the function can change the value of the pointer that is passed as an argument (as distinct from changing what the pointer points to). If you then supply an array (rather than a pointer) as the function's argument, the function definition would suggest that the function could change the value of the array as if it were a pointer, which is impossible. Thus the language requires that you declare that the pointer is const, so that the function won't be changing it. An array can then legally be converted to a pointer when passed as a function argument. -- John Carson [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Jerry Guest Posted: Sat Oct 29, 2005 7:45 pm    Post subject: Re: How to cast (char *) to (char *)& for a "call by referen Stefan, thanks for your reply. But my goal it to learn how to cast the PathA correctly. If it's not castable, why? [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top jeff_d_harper@hotmail.com Guest Posted: Sat Oct 29, 2005 7:47 pm    Post subject: Re: How to cast (char *) to (char *)& for a "call by referen The compiler complains because PathA is not a char pointer. It is a char array. PathA can be used in many places where a char * type is expected. However, one major difference between PathA and a normal char * is that PathA cannot be modified to point to a different address. Since foo takes a reference to a pointer, it could modify the caller's pointer. Since PathA cannot be modified, the compiler must complain. Your example is similar to this: void foo(int & x); main() { foo(3); // <- the compiler will complain here because // the literal 3 cannot be modified. However, // foo is allowed to change the int passed by the // caller. } You can make your code compile by changing the declaration of PathA to make it a char *: char *PathA = "MyPath"; It is a little bit unusual (though not unheard of) to pass references to pointers. Instead of making changes in main, you might be able to change the declaration of foo to make it more usable. Unless foo needs to change the caller's pointer, it's declaration could be changed to accept a char *: void foo(char * Path) ; [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Niklas Matthies Guest Posted: Sun Oct 30, 2005 3:32 am    Post subject: Re: How to cast (char *) to (char *)& for a "call by referen On 2005-10-29 19:45, Jerry wrote: Quote: Stefan, thanks for your reply. But my goal it to learn how to cast the PathA correctly. If it's not castable, why? It is castable, that is not the problem. The problem is that casting only creates a temporary object (of type 'char *'), and C++ doesn't allow temporary objects to be bound to non-const references. It will work if you change foo() to 'void foo(char * const & path)'. -- Niklas Matthies [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Carlos Moreno Guest Posted: Sun Oct 30, 2005 3:02 pm    Post subject: Re: How to cast (char *) to (char *)& for a "call by referen Jerry wrote: Quote: Stefan, thanks for your reply. But my goal it to learn how to cast the PathA correctly. If it's not castable, why? Jerry, Be very careful with this. Usually, when the compiler does not allow you to do something, it is because that something is not possible, or it is wrong. The fact that you are passing a char * by reference *most likely* means that you need to modify the pointer inside the function (and have that change modified in the client code's pointer that was passed to the function). If that is not the case, then your function is poorly designed, and you should not pass the parameter by reference. Now, if you do want to modify the pointer, and thus need to pass by reference, then you have to have *an actual pointer* to refer to, and not a temporary thing that is the outcome of an implicit conversion and that does not physically exist in the data of your program. You requrie something like a variable of type 'char *', so that when you refer to it and modify it, the operation is meaningful. A char array *is not* a pointer; you can not have a reference-to- pointer refer to this, because this is an object that is physically different from a pointer. If you pass by value (the value of the pointer), then everything is ok, since the implicit conversion from array to pointer allows the compiler to temporarily create a pointer, by taking the address of the first element of the array, and pass that value (that memory address) to the function. Same thing happens if you pass by reference-to-const, since that means that the function is not going to modify the referred pointer, so the compiler is happy with passing a reference to the temporary "ficticious" pointer. Again: the important thing is, make an effort to understand what's really happening in your program, instead of making an effort in learning how to *force* the compiler to do something that is wrong and that will lead you to no good. HTH, Carlos -- [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First         C++Talk.NET Forum Index -> C++ Language (Moderated) All times are GMT Goto page 1, 2  Next Page 1 of 2   Jump to: Select a forum C/C++----------------C LanguageC Language (Moderated)C++ Language (Moderated)C++ language (comp.lang.c++)C++ language, library and standards Non-English C/C++ Newsgroups----------------C++ (German)C++ (French)  You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot vote in polls in this forum Powered by phpBB © 2001, 2006 phpBB Group SEO toolkit © 2004-2006 webmedic.