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  4. A Simple C++ Question.

A Simple C++ Question.

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  • S Offline
    S Offline
    Sivaraman Dhamodharan
    wrote on last edited by
    #1

    Have a look at the below code:

    #include "stdafx.h"
    #include "conio.h"

    int _tmain(int argc, _TCHAR* argv[])
    {
    //Snippet 1
    char name[20] = "Some Name";
    printf("%s\n", name);
    name[1] = 'a';
    printf("%s\n", name);

    //Snippet 2
    char \* name2 = "Some Name";
    printf("%s\\n", name2);
    name2\[1\] = 'a';
    printf("%s\\n", name2);
    
    getch();
    return 0;
    

    }

    Why we get Runtime error when we modify name2[1] = 'a';?

    Programming Article

    D A M 3 Replies Last reply
    0
    • S Sivaraman Dhamodharan

      Have a look at the below code:

      #include "stdafx.h"
      #include "conio.h"

      int _tmain(int argc, _TCHAR* argv[])
      {
      //Snippet 1
      char name[20] = "Some Name";
      printf("%s\n", name);
      name[1] = 'a';
      printf("%s\n", name);

      //Snippet 2
      char \* name2 = "Some Name";
      printf("%s\\n", name2);
      name2\[1\] = 'a';
      printf("%s\\n", name2);
      
      getch();
      return 0;
      

      }

      Why we get Runtime error when we modify name2[1] = 'a';?

      Programming Article

      D Offline
      D Offline
      David Crow
      wrote on last edited by
      #2

      It all boils down to the difference between an array and a pointer. For name, you have set aside room for 20 characters (that are initialized to "Some Name"). For name2, you are pointing to a static piece of memory. That memory cannot be changed.

      "One man's wage rise is another man's price increase." - Harold Wilson

      "Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons

      "Show me a community that obeys the Ten Commandments and I'll show you a less crowded prison system." - Anonymous

      S 1 Reply Last reply
      0
      • S Sivaraman Dhamodharan

        Have a look at the below code:

        #include "stdafx.h"
        #include "conio.h"

        int _tmain(int argc, _TCHAR* argv[])
        {
        //Snippet 1
        char name[20] = "Some Name";
        printf("%s\n", name);
        name[1] = 'a';
        printf("%s\n", name);

        //Snippet 2
        char \* name2 = "Some Name";
        printf("%s\\n", name2);
        name2\[1\] = 'a';
        printf("%s\\n", name2);
        
        getch();
        return 0;
        

        }

        Why we get Runtime error when we modify name2[1] = 'a';?

        Programming Article

        A Offline
        A Offline
        Albert Holguin
        wrote on last edited by
        #3

        In this case... your definition of name2 is what is referred to as a string literal and points to data that is considered constant. See here[^].

        1 Reply Last reply
        0
        • S Sivaraman Dhamodharan

          Have a look at the below code:

          #include "stdafx.h"
          #include "conio.h"

          int _tmain(int argc, _TCHAR* argv[])
          {
          //Snippet 1
          char name[20] = "Some Name";
          printf("%s\n", name);
          name[1] = 'a';
          printf("%s\n", name);

          //Snippet 2
          char \* name2 = "Some Name";
          printf("%s\\n", name2);
          name2\[1\] = 'a';
          printf("%s\\n", name2);
          
          getch();
          return 0;
          

          }

          Why we get Runtime error when we modify name2[1] = 'a';?

          Programming Article

          M Offline
          M Offline
          Maximilien
          wrote on last edited by
          #4

          You are trying to modify a const char array (pointer). This is not C++. if you were writing proper C++, you would not get those problems (or a lot less likely).

          std::string name = "some name";
          std::cout << name << std::endl;

          std::string name2 = "some name";
          std::cout << name2 << std::endl;

          name2[1] = 'a'; // need to check that the index is within the string size.
          std::cout << name2 << std::endl;

          Watched code never compiles.

          C A 2 Replies Last reply
          0
          • M Maximilien

            You are trying to modify a const char array (pointer). This is not C++. if you were writing proper C++, you would not get those problems (or a lot less likely).

            std::string name = "some name";
            std::cout << name << std::endl;

            std::string name2 = "some name";
            std::cout << name2 << std::endl;

            name2[1] = 'a'; // need to check that the index is within the string size.
            std::cout << name2 << std::endl;

            Watched code never compiles.

            C Offline
            C Offline
            Chuck OToole
            wrote on last edited by
            #5

            Technically, to pick nits, it *is* C++ as C++ is a superset of C. What you are really saying is that it isn't "Object Oriented Programming" which is true, but it is still within the definition of C++.

            J A 2 Replies Last reply
            0
            • M Maximilien

              You are trying to modify a const char array (pointer). This is not C++. if you were writing proper C++, you would not get those problems (or a lot less likely).

              std::string name = "some name";
              std::cout << name << std::endl;

              std::string name2 = "some name";
              std::cout << name2 << std::endl;

              name2[1] = 'a'; // need to check that the index is within the string size.
              std::cout << name2 << std::endl;

              Watched code never compiles.

              A Offline
              A Offline
              Albert Holguin
              wrote on last edited by
              #6

              Agree with Chuck... technically it is C++.

              1 Reply Last reply
              0
              • C Chuck OToole

                Technically, to pick nits, it *is* C++ as C++ is a superset of C. What you are really saying is that it isn't "Object Oriented Programming" which is true, but it is still within the definition of C++.

                J Offline
                J Offline
                jschell
                wrote on last edited by
                #7

                Chuck O'Toole wrote:

                Technically, to pick nits, it *is* C++ as C++ is a superset of C

                Does the newest ANSI C++ standard incorporate the latest ANSI C standard as a subset? Last I heard that was not the case

                C 1 Reply Last reply
                0
                • C Chuck OToole

                  Technically, to pick nits, it *is* C++ as C++ is a superset of C. What you are really saying is that it isn't "Object Oriented Programming" which is true, but it is still within the definition of C++.

                  A Offline
                  A Offline
                  Aescleal
                  wrote on last edited by
                  #8

                  C++98 is not a superset of C90 - especially when you consider const and what happens when you cast away const. C allows modification of literals (string and constants) through pointers while C++ doesn't or rather considers it undefined behaviour.

                  C 1 Reply Last reply
                  0
                  • J jschell

                    Chuck O'Toole wrote:

                    Technically, to pick nits, it *is* C++ as C++ is a superset of C

                    Does the newest ANSI C++ standard incorporate the latest ANSI C standard as a subset? Last I heard that was not the case

                    C Offline
                    C Offline
                    Chuck OToole
                    wrote on last edited by
                    #9

                    The point is, C++ did *not* decommit the "char" data type nor pointers. std::string is *not* the only way to do string manipulation in C++. It may be the perferred way for some people but it's not the only way.

                    J 1 Reply Last reply
                    0
                    • A Aescleal

                      C++98 is not a superset of C90 - especially when you consider const and what happens when you cast away const. C allows modification of literals (string and constants) through pointers while C++ doesn't or rather considers it undefined behaviour.

                      C Offline
                      C Offline
                      Chuck OToole
                      wrote on last edited by
                      #10

                      The point is, C++ did *not* decommit the "char" data type nor pointers. std::string is *not* the only way to do string manipulation in C++. It may be the perferred way for some people but it's not the only way.

                      1 Reply Last reply
                      0
                      • C Chuck OToole

                        The point is, C++ did *not* decommit the "char" data type nor pointers. std::string is *not* the only way to do string manipulation in C++. It may be the perferred way for some people but it's not the only way.

                        J Offline
                        J Offline
                        jschell
                        wrote on last edited by
                        #11

                        Chuck O'Toole wrote:

                        The point is, C++ did *not* decommit the "char" data type nor pointer

                        I was responding to what appeared to be a general comment about the language in general and in its entirety and not just one small part. And that general comment, at this time (new standards), is wrong.

                        C 1 Reply Last reply
                        0
                        • J jschell

                          Chuck O'Toole wrote:

                          The point is, C++ did *not* decommit the "char" data type nor pointer

                          I was responding to what appeared to be a general comment about the language in general and in its entirety and not just one small part. And that general comment, at this time (new standards), is wrong.

                          C Offline
                          C Offline
                          Chuck OToole
                          wrote on last edited by
                          #12

                          Context dude, gotta read comments in the context of the thread

                          J 1 Reply Last reply
                          0
                          • C Chuck OToole

                            Context dude, gotta read comments in the context of the thread

                            J Offline
                            J Offline
                            jschell
                            wrote on last edited by
                            #13

                            Chuck O'Toole wrote:

                            Context dude, gotta read comments in the context of the thread

                            And respond in that context. Since another poster read your response in way similar to the way I read it, it suggests that your response is the one at fault.

                            1 Reply Last reply
                            0
                            • D David Crow

                              It all boils down to the difference between an array and a pointer. For name, you have set aside room for 20 characters (that are initialized to "Some Name"). For name2, you are pointing to a static piece of memory. That memory cannot be changed.

                              "One man's wage rise is another man's price increase." - Harold Wilson

                              "Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons

                              "Show me a community that obeys the Ten Commandments and I'll show you a less crowded prison system." - Anonymous

                              S Offline
                              S Offline
                              Sivaraman Dhamodharan
                              wrote on last edited by
                              #14

                              Thanks Crow. That is constant piece of memory.

                              Programming Article

                              1 Reply Last reply
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