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  3. function overload problem:(char) and (char*)

function overload problem:(char) and (char*)

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  • C CP Visitor

    jan larsen wrote: For example, when a VB programmer comes to my house, they may say 'does your pool need cleaning, sir ?' " This not correct English, no?

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    jan larsen
    wrote on last edited by
    #9

    What did you expect from VB programmers :-) "After all it's just text at the end of the day. - Colin Davies "For example, when a VB programmer comes to my house, they may say 'does your pool need cleaning, sir ?' " - Christian Graus

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    • D Daniel Turini

      Did you try calling fun(0*)? It's obvious that you're passing a single char (the zero) to the function, so the compiler will call the fun(char) overload. Just put an asterisk and everything should work. Ok, I reckon, this is horrible, but it's not worse than "the compiler is having fun()" I see dead pixels Yes, even I am blogging now!

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      jan larsen
      wrote on last edited by
      #10

      Daniel Turini wrote: Did you try calling fun(0*)? Oh that one's obvious, as long as he remember to add the compiler hint at the end of the code like this:

      int main()
      {
      fun(0*);
      }
      *: It's (char) 0, not (void *) 0, now go on...

      "After all it's just text at the end of the day. - Colin Davies "For example, when a VB programmer comes to my house, they may say 'does your pool need cleaning, sir ?' " - Christian Graus

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      • D Daniel Turini

        Did you try calling fun(0*)? It's obvious that you're passing a single char (the zero) to the function, so the compiler will call the fun(char) overload. Just put an asterisk and everything should work. Ok, I reckon, this is horrible, but it's not worse than "the compiler is having fun()" I see dead pixels Yes, even I am blogging now!

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        bobi_zcl
        wrote on last edited by
        #11

        I call fun(0),the compiler display message "ambiguous call...",when I call fun(1)(passing a value which is not zero),the compiler call fun(char),why?? your friend:bobi

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        • D Daniel Turini

          Did you try calling fun(0*)? It's obvious that you're passing a single char (the zero) to the function, so the compiler will call the fun(char) overload. Just put an asterisk and everything should work. Ok, I reckon, this is horrible, but it's not worse than "the compiler is having fun()" I see dead pixels Yes, even I am blogging now!

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          Aamir Butt
          wrote on last edited by
          #12

          what about fun(*0). This will lead to the first address in memory and then will have some fun with it :) "Aerodynamically, the bumble bee shouldn't be able to fly, but the bumble bee doesn't know it so it goes on flying anyway." - Mary Kay Ash My Articles

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          • B bobi_zcl

            I still havn't understand what you said,Can you explain? your friend:bobi

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            jan larsen
            wrote on last edited by
            #13

            It was a joke, programming questions are to be posted in the programming fora. This is the lounge, people asking programming questions in the lounge will be put in the comfy chair. I will probably regret this, but here is the real answer: When you pass the literal 0 to your function, the compiler will have no way of knowing if it is (char) 0 or (char *) 0. You will have to explicitly type cast it. In the Visual C++ forum you would have had the opportunity to get a better answer from eg. Christian Graus involving standard C++ type casts, I can only offer C style casts :-) : fun((char *) 0) or fun((char) 0). "After all it's just text at the end of the day. - Colin Davies "For example, when a VB programmer comes to my house, they may say 'does your pool need cleaning, sir ?' " - Christian Graus

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            • B bobi_zcl

              I call fun(0),the compiler display message "ambiguous call...",when I call fun(1)(passing a value which is not zero),the compiler call fun(char),why?? your friend:bobi

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              V 0
              wrote on last edited by
              #14

              on top of this page on the right you see "Message Boards", if you click on that you will get: OR a dropdown menu OR it's a link If you get the dropdownmenu click on the Visual C++ and you will be taken to the C++ forum, else you will get a bunch of links including Visual C++, which will also take you to the C++ forum. When you get there (the Visual C++ forum) kindly ask your question again and it will probably be answered without jokes. good luck. No hurries, no worries.

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              • B bobi_zcl

                I call fun(0),the compiler display message "ambiguous call...",when I call fun(1)(passing a value which is not zero),the compiler call fun(char),why?? your friend:bobi

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                Jorgen Sigvardsson
                wrote on last edited by
                #15

                Because 0 is a valid pointer constant in C++, 1 isn't. Since 1 isn't a valid pointer constant, the compiler will ignore any funs with taking pointers as argument. Anyway, NO PROGRAMMING QUESTIONS IN THE LOUNGE! Next time, we'll demand your first born. -- My name in Katakana is ヨルゲン. My name in German is Jörgen. My name in Mandarin/Kanji is 乔尔根 西格瓦德森. My name in Korean is 요르겐. I blog too now[^]

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                • J Jorgen Sigvardsson

                  Because 0 is a valid pointer constant in C++, 1 isn't. Since 1 isn't a valid pointer constant, the compiler will ignore any funs with taking pointers as argument. Anyway, NO PROGRAMMING QUESTIONS IN THE LOUNGE! Next time, we'll demand your first born. -- My name in Katakana is ヨルゲン. My name in German is Jörgen. My name in Mandarin/Kanji is 乔尔根 西格瓦德森. My name in Korean is 요르겐. I blog too now[^]

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                  leppie
                  wrote on last edited by
                  #16

                  Jörgen Sigvardsson wrote: Next time, we'll demand your first born. Your really want that? :p xacc-ide 0.0.15 now with C#, MSIL, C, XML, ASP.NET, Nemerle, MyXaml and HLSL coloring - Screenshots

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                  • J jan larsen

                    It was a joke, programming questions are to be posted in the programming fora. This is the lounge, people asking programming questions in the lounge will be put in the comfy chair. I will probably regret this, but here is the real answer: When you pass the literal 0 to your function, the compiler will have no way of knowing if it is (char) 0 or (char *) 0. You will have to explicitly type cast it. In the Visual C++ forum you would have had the opportunity to get a better answer from eg. Christian Graus involving standard C++ type casts, I can only offer C style casts :-) : fun((char *) 0) or fun((char) 0). "After all it's just text at the end of the day. - Colin Davies "For example, when a VB programmer comes to my house, they may say 'does your pool need cleaning, sir ?' " - Christian Graus

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                    KaRl
                    wrote on last edited by
                    #17

                    jan larsen wrote: I will probably regret this, but here is the real answer: Shame on you! You can't stop yourself to help people, can you, you bloody socialist evildoer :-D


                    Fold With Us! Sie wollen mein Herz am rechten Fleck, doch Seh' ich dann nach unten weg Da schlägt es links!

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                    • L leppie

                      Jörgen Sigvardsson wrote: Next time, we'll demand your first born. Your really want that? :p xacc-ide 0.0.15 now with C#, MSIL, C, XML, ASP.NET, Nemerle, MyXaml and HLSL coloring - Screenshots

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                      Jorgen Sigvardsson
                      wrote on last edited by
                      #18

                      No, but he will probably want his first born more than I do. That's sort of the point. ;P -- My name in Katakana is ヨルゲン. My name in German is Jörgen. My name in Mandarin/Kanji is 乔尔根 西格瓦德森. My name in Korean is 요르겐. I blog too now[^]

                      1 Reply Last reply
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                      • L leppie

                        Jörgen Sigvardsson wrote: Next time, we'll demand your first born. Your really want that? :p xacc-ide 0.0.15 now with C#, MSIL, C, XML, ASP.NET, Nemerle, MyXaml and HLSL coloring - Screenshots

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                        KaRl
                        wrote on last edited by
                        #19

                        Please don't ask![^] :-D


                        Fold With Us! Sie wollen mein Herz am rechten Fleck, doch Seh' ich dann nach unten weg Da schlägt es links!

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                        • J jan larsen

                          It was a joke, programming questions are to be posted in the programming fora. This is the lounge, people asking programming questions in the lounge will be put in the comfy chair. I will probably regret this, but here is the real answer: When you pass the literal 0 to your function, the compiler will have no way of knowing if it is (char) 0 or (char *) 0. You will have to explicitly type cast it. In the Visual C++ forum you would have had the opportunity to get a better answer from eg. Christian Graus involving standard C++ type casts, I can only offer C style casts :-) : fun((char *) 0) or fun((char) 0). "After all it's just text at the end of the day. - Colin Davies "For example, when a VB programmer comes to my house, they may say 'does your pool need cleaning, sir ?' " - Christian Graus

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                          peterchen
                          wrote on last edited by
                          #20

                          sorry, nothing personal. :rolleyes:


                          Pandoras Gift #44: Hope. The one that keeps you on suffering.
                          aber.. "Wie gesagt, der Scheiss is' Therapie"
                          boost your code || Fold With Us! || sighist | doxygen

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                          • J jan larsen

                            It was a joke, programming questions are to be posted in the programming fora. This is the lounge, people asking programming questions in the lounge will be put in the comfy chair. I will probably regret this, but here is the real answer: When you pass the literal 0 to your function, the compiler will have no way of knowing if it is (char) 0 or (char *) 0. You will have to explicitly type cast it. In the Visual C++ forum you would have had the opportunity to get a better answer from eg. Christian Graus involving standard C++ type casts, I can only offer C style casts :-) : fun((char *) 0) or fun((char) 0). "After all it's just text at the end of the day. - Colin Davies "For example, when a VB programmer comes to my house, they may say 'does your pool need cleaning, sir ?' " - Christian Graus

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                            bobi_zcl
                            wrote on last edited by
                            #21

                            I can understand.but I define another two:fun(int) and fun(int*),then I call fun(0),the compiler will compile successfully, and fun(0) will select fun(int) instead of fun(int*).why? thank your suggestion,I will go to the visual C++ forum. your friend:bobi

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