Pointers and arrays.

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  • raashid bhatt
    #1

    Pointers and arrays.

    #include <string.h>
    #include <stdio.h>
    #include <windows.h>


    int main(int argc, char *argv[])
    {

    char s[10] = "welcome";

    char *p = s;
    char **prt = p; //error happens here
    return 0;
    }


    Tags: None
    • Mike Wahler
      #2
      Re: Pointers and arrays.


      "raashid bhatt" <raashidbhatt@g mail.comwrote in message
      news:7c9c5378-ffd0-485b-8cd4-66e0f5f5e500@t1 g2000pra.google groups.com...
      #include <string.h>
      #include <stdio.h>
      #include <windows.h>
      <windows.his not a standard C header. Please
      omit such from code posted here (it has nothing
      to do with your question anyway).
      >
      >
      int main(int argc, char *argv[])
      {
      >
      char s[10] = "welcome";
      >
      char *p = s;
      char **prt = p; //error happens here
      That's because you're trying to assign
      an incompatible type. Write:

      char **prt = &p;

      return 0;
      }

      I've shown you correct syntax. But what
      exactly are you trying to do?

      -Mike


        Comment

        • raashid bhatt
          #3
          Re: Pointers and arrays.

          On Aug 19, 6:54 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
          "raashid bhatt" <raashidbh...@g mail.comwrote in message
          >
          news:7c9c5378-ffd0-485b-8cd4-66e0f5f5e500@t1 g2000pra.google groups.com...
          >
          #include <string.h>
          #include <stdio.h>
          #include <windows.h>
          >
          <windows.his not a standard C header. Please
          omit such from code posted here (it has nothing
          to do with your question anyway).
          >
          >
          >
          int main(int argc, char *argv[])
          {
          >
          char s[10] = "welcome";
          >
          char *p = s;
          char **prt = p; //error happens here
          >
          That's because you're trying to assign
          an incompatible type.  Write:
          >
          char **prt = &p;
          >
          return 0;
          }
          >
          I've shown you correct syntax.  But what
          exactly are you trying to do?
          >
          -Mike
          this is wrong man why do i need &p as p consists the address

            Comment

            • Nick Keighley
              #4
              Re: Pointers and arrays.

              On 19 Aug, 15:01, raashid bhatt <raashidbh...@g mail.comwrote:
              On Aug 19, 6:54 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
              >
              >
              >
              >
              >
              "raashid bhatt" <raashidbh...@g mail.comwrote in message
              >
              news:7c9c5378-ffd0-485b-8cd4-66e0f5f5e500@t1 g2000pra.google groups.com...
              >
              #include <string.h>
              #include <stdio.h>
              #include <windows.h>
              >
              <windows.his not a standard C header. Please
              omit such from code posted here (it has nothing
              to do with your question anyway).
              >
              int main(int argc, char *argv[])
              {
              >
              char s[10] = "welcome";
              >
              char *p = s;
              char **prt = p; //error happens here
              >
              That's because you're trying to assign
              an incompatible type.  Write:
              >
              char **prt = &p;
              >
              return 0;
              }
              >
              I've shown you correct syntax.  But what
              exactly are you trying to do?
              >
              -Mike

              this is wrong man why do i need &p as p consists the address
              no it *isn't* wrong. Consider the types, p has type char* (ptr-to-
              char)
              prt has type char** (ptr-to -ptr-to-char). They are not the same.
              But &p is the address of a ptr-to-char so it has type ptr-to-ptr-to-
              char.
              So that assignment is ok.

              Get a good text book. And read it.


              --
              Nick Keighley




                Comment

                • Jens Thoms Toerring
                  #5
                  Re: Pointers and arrays.

                  raashid bhatt <raashidbhatt@g mail.comwrote:
                  #include <string.h>
                  #include <stdio.h>
                  #include <windows.h>
                  Do you realize that this is a non-standard header (and that
                  rather likely it's not needed)?
                  int main(int argc, char *argv[])
                  {
                  char s[10] = "welcome";
                  char *p = s;
                  This works because in this context ('s' is used as a value)
                  it is automatically converted to a pointer to the first
                  char of 's'. So you have a char pointer on the left and
                  n the right hand side of the assignment.
                  char **prt = p; //error happens here
                  But here you try to assign the value of a char pointer to a
                  pointer to pointer to char, so you have different types on
                  the left and right hand side. Did you try

                  char **prt = &p;
                  return 0;
                  }
                  Regards, Jens
                  --
                  \ Jens Thoms Toerring ___ jt@toerring.de
                  \______________ ____________ http://toerring.de

                    Comment

                    • santosh
                      #6
                      Re: Pointers and arrays.

                      raashid bhatt wrote:
                      #include <string.h>
                      #include <stdio.h>
                      #include <windows.h>
                      >
                      >
                      int main(int argc, char *argv[])
                      {
                      >
                      char s[10] = "welcome";
                      >
                      char *p = s;
                      char **prt = p; //error happens here
                      prt is of type char** or "pointer to pointer to char". The
                      expression 'p' on the RHS of the assignment resolves to a value of type
                      char* which is not a compatible type for prt. You need to store the
                      *address* of char* object, which you can derive in this case by
                      applying the address-of operator to 'p'. Hence:

                      char **prt = &p;

                      is correct, as would be:

                      char **prt = &s;

                      or

                      char **prt = &p[5];

                      or even

                      char **prt = &p[10];

                      though the last one can be computed and stored but not deferenced.

                      <snip>

                        Comment

                        • santosh
                          #7
                          Re: Pointers and arrays.

                          raashid bhatt wrote:
                          On Aug 19, 6:54 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
                          >"raashid bhatt" <raashidbh...@g mail.comwrote in message
                          >>
                          >>
                          news:7c9c5378-ffd0-485b-8cd4-66e0f5f5e500@t1 g2000pra.google groups.com...
                          >>
                          #include <string.h>
                          #include <stdio.h>
                          #include <windows.h>
                          >>
                          ><windows.his not a standard C header. Please
                          >omit such from code posted here (it has nothing
                          >to do with your question anyway).
                          >>
                          >>
                          >>
                          int main(int argc, char *argv[])
                          {
                          >>
                          char s[10] = "welcome";
                          >>
                          char *p = s;
                          char **prt = p; //error happens here
                          >>
                          >That's because you're trying to assign
                          >an incompatible type.  Write:
                          >>
                          >char **prt = &p;
                          >>
                          return 0;
                          }
                          >>
                          >I've shown you correct syntax.  But what
                          >exactly are you trying to do?
                          this is wrong man why do i need &p as p consists the address
                          The expression 'p' evaluates to the *value* of 'p', *not* to it's
                          address. Consider:

                          int i = 5;
                          int j = i;

                          Now what does the expression on the RHS of the second declaration
                          evaluate to? The address or the value of 'i'? Why do you think pointers
                          are any different?

                          int i = 5;
                          int *ip = &i;
                          int **ipp = &ip;

                          In an expression context 'i' will evaluate to the value stored in the
                          object that it designates. To initialise ip though we need the address
                          of i, so we apply the address-of operator to get it. The expression ip
                          will evaluate to the value that it contains, i.e., the address of the
                          int object to which it has been set to point to, or the address NULL,
                          or an indeterminate value. To get the *address* of ip we apply the
                          address-of operator on it, exactly as was the case for i.

                          As Eric said elsewhere, reading your textbook might be a better approach
                          than asking multitudes of elementary questions. At the very least go
                          over the CLC FAQ whose link I gave you earlier.

                            Comment

                            • Mike Wahler
                              #8
                              Re: Pointers and arrays.

                              "raashid bhatt" <raashidbhatt@g mail.comwrote in message
                              news:ac09bd76-aee9-4940-9858-2d37c42c2b4c@z6 g2000pre.google groups.com...
                              On Aug 19, 6:54 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
                              "raashid bhatt" <raashidbh...@g mail.comwrote in message
                              >
                              news:7c9c5378-ffd0-485b-8cd4-66e0f5f5e500@t1 g2000pra.google groups.com...
                              char s[10] = "welcome";
                              >>
                              char *p = s;
                              char **prt = p; //error happens here
                              >>
                              >That's because you're trying to assign
                              >an incompatible type. Write:
                              >>
                              >char **prt = &p;
                              >
                              return 0;
                              }
                              >>
                              >I've shown you correct syntax. But what
                              >exactly are you trying to do?
                              >>
                              >-Mike
                              this is wrong man why do i need &p as p consists the address
                              What I wrote is not wrong, it's correct. My assertion is
                              supported by Kernighan & Ritchie (inventors of the C language),
                              The international standard for the C language, and 20 years of
                              experience using the C language.

                              What supports your claim that what I wrote is wrong?

                              -Mike


                                Comment

                                • raashid bhatt
                                  #9
                                  Re: Pointers and arrays.

                                  On Aug 19, 8:57 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
                                  "raashid bhatt" <raashidbh...@g mail.comwrote in message
                                  >
                                  news:ac09bd76-aee9-4940-9858-2d37c42c2b4c@z6 g2000pre.google groups.com...
                                  On Aug 19, 6:54 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
                                  >
                                  >
                                  >
                                  "raashid bhatt" <raashidbh...@g mail.comwrote in message
                                  >
                                  news:7c9c5378-ffd0-485b-8cd4-66e0f5f5e500@t1 g2000pra.google groups.com...
                                  char s[10] = "welcome";
                                  >
                                  char *p = s;
                                  char **prt = p; //error happens here
                                  >
                                  That's because you're trying to assign
                                  an incompatible type. Write:
                                  >
                                  char **prt = &p;
                                  >
                                  return 0;
                                  }
                                  >
                                  I've shown you correct syntax. But what
                                  exactly are you trying to do?
                                  >
                                  -Mike
                                  this is wrong man why do i need &p as p consists the address
                                  >
                                  What I wrote is not wrong, it's correct.  My assertion is
                                  supported by Kernighan & Ritchie (inventors of the C language),
                                  The international standard for the C language, and 20 years of
                                  experience using the C language.
                                  >
                                  What supports your claim that what I wrote is wrong?
                                  >
                                  -Mike
                                  This is wrong man correct ur thing

                                    Comment

                                    • Joachim Schmitz
                                      #10
                                      Re: Pointers and arrays.

                                      raashid bhatt wrote:
                                      On Aug 19, 8:57 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
                                      >"raashid bhatt" <raashidbh...@g mail.comwrote in message
                                      >>
                                      >news:ac09bd7 6-aee9-4940-9858-2d37c42c2b4c@z6 g2000pre.google groups.com...
                                      >On Aug 19, 6:54 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
                                      >>
                                      >>
                                      >>
                                      >>"raashid bhatt" <raashidbh...@g mail.comwrote in message
                                      >>
                                      >>news:7c9c53 78-ffd0-485b-8cd4-66e0f5f5e500@t1 g2000pra.google groups.com...
                                      >>>>char s[10] = "welcome";
                                      >>
                                      >>>>char *p = s;
                                      >>>>char **prt = p; //error happens here
                                      >>
                                      >>>That's because you're trying to assign
                                      >>>an incompatible type. Write:
                                      >>
                                      >>>char **prt = &p;
                                      >>
                                      >>>>return 0;
                                      >>>>}
                                      >>
                                      >>>I've shown you correct syntax. But what
                                      >>>exactly are you trying to do?
                                      >>
                                      >>>-Mike
                                      >>this is wrong man why do i need &p as p consists the address
                                      >>
                                      >What I wrote is not wrong, it's correct. My assertion is
                                      >supported by Kernighan & Ritchie (inventors of the C language),
                                      >The international standard for the C language, and 20 years of
                                      >experience using the C language.
                                      >>
                                      >What supports your claim that what I wrote is wrong?
                                      >>
                                      >-Mike
                                      >
                                      This is wrong man correct ur thing
                                      If you know better, why did you ask in the first place?

                                      Bye, Jojo


                                        Comment

                                        • santosh
                                          #11
                                          Re: Pointers and arrays.

                                          raashid bhatt wrote:
                                          On Aug 19, 8:57 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
                                          >"raashid bhatt" <raashidbh...@g mail.comwrote in message
                                          >>
                                          >>
                                          news:ac09bd76-aee9-4940-9858-2d37c42c2b4c@z6 g2000pre.google groups.com...
                                          >On Aug 19, 6:54 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
                                          >>
                                          >>
                                          >>
                                          "raashid bhatt" <raashidbh...@g mail.comwrote in message
                                          >>
                                          >>
                                          >news:7c9c537 8-ffd0-485b-8cd4-66e0f5f5e500@t1 g2000pra.google groups.com...
                                          char s[10] = "welcome";
                                          >>
                                          char *p = s;
                                          char **prt = p; //error happens here
                                          >>
                                          >That's because you're trying to assign
                                          >an incompatible type. Write:
                                          >>
                                          >char **prt = &p;
                                          >>
                                          return 0;
                                          }
                                          >>
                                          >I've shown you correct syntax. But what
                                          >exactly are you trying to do?
                                          >>
                                          >-Mike
                                          this is wrong man why do i need &p as p consists the address
                                          >>
                                          >What I wrote is not wrong, it's correct.  My assertion is
                                          >supported by Kernighan & Ritchie (inventors of the C language),
                                          >The international standard for the C language, and 20 years of
                                          >experience using the C language.
                                          >>
                                          >What supports your claim that what I wrote is wrong?
                                          >>
                                          >-Mike
                                          >
                                          This is wrong man correct ur thing
                                          Right. I think we can file this one as another troll.

                                            Comment

                                            • Nick Keighley
                                              #12
                                              Re: Pointers and arrays.

                                              On 19 Aug, 17:05, raashid bhatt <raashidbh...@g mail.comwrote:
                                              On Aug 19, 8:57 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
                                              "raashid bhatt" <raashidbh...@g mail.comwrote in message
                                              news:ac09bd76-aee9-4940-9858-2d37c42c2b4c@z6 g2000pre.google groups.com...
                                              On Aug 19, 6:54 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
                                              "raashid bhatt" <raashidbh...@g mail.comwrote in message
                                              >news:7c9c537 8-ffd0-485b-8cd4-66e0f5f5e500@t1 g2000pra.google groups.com....
                                              char s[10] = "welcome";
                                              >
                                              char *p = s;
                                              char **prt = p; //error happens here
                                              >
                                              >That's because you're trying to assign
                                              >an incompatible type. Write:
                                              >
                                              >char **prt = &p;
                                              >
                                              return 0;
                                              }
                                              >
                                              >I've shown you correct syntax. But what
                                              >exactly are you trying to do?
                                              >
                                              this is wrong man why do i need &p as p consists the address
                                              >
                                              What I wrote is not wrong, it's correct.  My assertion is
                                              supported by Kernighan & Ritchie (inventors of the C language),
                                              The international standard for the C language, and 20 years of
                                              experience using the C language.
                                              >
                                              What supports your claim that what I wrote is wrong?
                                              >
                                              This is wrong man correct ur thing
                                              what does that mean? Write in english.

                                              plonk

                                              --
                                              Nick Keighley

                                                Comment

                                                • John Bode
                                                  #13
                                                  Re: Pointers and arrays.

                                                  On Aug 19, 11:05 am, raashid bhatt <raashidbh...@g mail.comwrote:
                                                  On Aug 19, 8:57 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
                                                  >
                                                  >
                                                  >
                                                  "raashid bhatt" <raashidbh...@g mail.comwrote in message
                                                  >
                                                  news:ac09bd76-aee9-4940-9858-2d37c42c2b4c@z6 g2000pre.google groups.com...
                                                  On Aug 19, 6:54 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
                                                  >
                                                  "raashid bhatt" <raashidbh...@g mail.comwrote in message
                                                  >
                                                  >news:7c9c537 8-ffd0-485b-8cd4-66e0f5f5e500@t1 g2000pra.google groups.com...
                                                  char s[10] = "welcome";
                                                  >
                                                  char *p = s;
                                                  char **prt = p; //error happens here
                                                  >
                                                  >That's because you're trying to assign
                                                  >an incompatible type. Write:
                                                  >
                                                  >char **prt = &p;
                                                  >
                                                  return 0;
                                                  }
                                                  >
                                                  >I've shown you correct syntax. But what
                                                  >exactly are you trying to do?
                                                  >
                                                  >-Mike
                                                  this is wrong man why do i need &p as p consists the address
                                                  >
                                                  What I wrote is not wrong, it's correct. My assertion is
                                                  supported by Kernighan & Ritchie (inventors of the C language),
                                                  The international standard for the C language, and 20 years of
                                                  experience using the C language.
                                                  >
                                                  What supports your claim that what I wrote is wrong?
                                                  >
                                                  -Mike
                                                  >
                                                  This is wrong man correct ur thing
                                                  It is NOT wrong; check your handy C reference manual or the FAQ. When
                                                  doing an assignment, the *types* of both the source and the target
                                                  have to be compatible. The *type* of p is char*, which is not the
                                                  same as the type of prt, which is char**.

                                                  prt is looking for a pointer-to-pointer-to-char. p is simply pointer-
                                                  to-char. However, &p evaluates to pointer-to-pointer-to-char. So the
                                                  answer, as several people have already pointed out, is to write

                                                  char **prt = &p;

                                                  If you still think this is wrong, then give up on programming in C;
                                                  you're just wasting your time otherwise.

                                                    Comment

                                                    • Anand Hariharan
                                                      #14
                                                      Re: Pointers and arrays.

                                                      On Tue, 19 Aug 2008 09:05:38 -0700, raashid bhatt <raashidbhatt@g mail.com>
                                                      wrote:
                                                      On Aug 19, 8:57 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
                                                      >"raashid bhatt" <raashidbh...@g mail.comwrote in message
                                                      >>
                                                      >news:ac09bd7 6-
                                                      aee9-4940-9858-2d37c42c2b4c@z6 g200...le groups.com...
                                                      >On Aug 19, 6:54 am, "Mike Wahler" <mkwah...@mkwah ler.netwrote:
                                                      >>
                                                      >>
                                                      >>
                                                      "raashid bhatt" <raashidbh...@g mail.comwrote in message
                                                      >>
                                                      >news:7c9c537 8-
                                                      ffd0-485b-8cd4-66e0f5f5e500@t1 g200...le groups.com...
                                                      char s[10] = "welcome";
                                                      >>
                                                      char *p = s;
                                                      char **prt = p; //error happens here
                                                      >>
                                                      >That's because you're trying to assign an incompatible type. Write:
                                                      >>
                                                      >char **prt = &p;
                                                      >>
                                                      (...)
                                                      this is wrong man why do i need &p as p consists the address
                                                      >>
                                                      Did you read Santosh's response to this question?

                                                      >What I wrote is not wrong, it's correct.  My assertion is supported by
                                                      >Kernighan & Ritchie (inventors of the C language), The international
                                                      >standard for the C language, and 20 years of experience using the C
                                                      >language.
                                                      >>
                                                      >What supports your claim that what I wrote is wrong?
                                                      >>
                                                      >-Mike
                                                      >
                                                      This is wrong man correct ur thing
                                                      And I suppose the compiler is wrong as well? Did you actually read
                                                      Mike's question?

                                                      - Anand

                                                        Comment

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