function pointer question

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  • MK
    #1

    function pointer question

    Hello,
    I have been trying to get a handle on the following piece of code.
    ----- BEGIN ----
    #include <stdio.h>

    int
    f2args(int a,int b)
    {
    printf("a=%d,b= %d\n",a,b);
    return 1;
    }

    typedef int (*func3args)(in t,int,int);

    int
    main(int argc,char **argv)
    {
    func3args fp = (func3args)f;
    fp(1,2,3);
    exit(0);
    }
    ----- END ----

    weird things:
    - I get no compiler warnings or errors (porbably because I used a cast
    (func)).
    - And when I run I expect the program to crash, but it runs fine.

    $ ./a.out
    a=1,b=2

    My questions are:
    - Is this portable and guranteed by the C standard?
    - Or is it the manifestation of the C calling convention (the
    default), hence not portable or
    implementation dependent.
    I dug aroung and saw a lot of information regarding 'function pointer
    compatibility', I guess I
    don't fully understand that.

    Thanks
    MK

    Tags: None
    • raxip
      #2
      Re: function pointer question

      I don't even see how that program would compile. Do you mean
      'func3args fp = (func3args)f2ar gs'?

      If you did, you should get a compiler error complaining about too many
      arguements.

        Comment

        • MK
          #3
          Re: function pointer question

          raxip wrote:[color=blue]
          > I don't even see how that program would compile. Do you mean
          > 'func3args fp = (func3args)f2ar gs'?
          >
          > If you did, you should get a compiler error complaining about too many
          > arguements.[/color]

          raxip:

          I am sorry something went awry with the cut and paste yes it was, here
          is the whole
          program and compile and run log (Linux x86, gcc 3.3.5).
          --- Begin xx.c ----
          $ cat xx.c
          #include <stdio.h>

          int
          func2args(int a,int b)
          {
          printf("a=%d,b= %d\n",a,b);
          return 1;
          }

          typedef int (*func3args)(in t,int,int);

          int
          main(int argc,char **argv)
          {
          func3args fp = (func3args)func 2args;
          fp(1,2,3);
          exit(0);
          }
          ------------------ End xx.c -------
          $ cc xx.c

          $ ./a.out
          a=1,b=2

          --------------------------
          Now, I understand that casting func2args probably is 'shutting' the
          compiler up. As I believe it should be legal to do that.

            Comment

            • Keith Thompson
              #4
              Re: function pointer question

              "MK" <Wavy2Gravy@gma il.com> writes:[color=blue]
              > Hello,
              > I have been trying to get a handle on the following piece of code.
              > ----- BEGIN ----
              > #include <stdio.h>
              >
              > int
              > f2args(int a,int b)
              > {
              > printf("a=%d,b= %d\n",a,b);
              > return 1;
              > }
              >
              > typedef int (*func3args)(in t,int,int);
              >
              > int
              > main(int argc,char **argv)
              > {
              > func3args fp = (func3args)f;
              > fp(1,2,3);
              > exit(0);
              > }
              > ----- END ----
              >
              > weird things:
              > - I get no compiler warnings or errors (porbably because I used a cast
              > (func)).[/color]

              When I compiled it, I got:

              tmp.c: In function `main':
              tmp.c:15: error: `f' undeclared (first use in this function)
              tmp.c:15: error: (Each undeclared identifier is reported only once
              tmp.c:15: error: for each function it appears in.)

              I presume your line
              func3args fp = (func3args)f;
              should be
              func3args fp = (func3args)f2ar gs;

              If you're going to post code, it's very important to post the *exact*
              code that you compiled. Copy-and-paste it, don't re-type it.
              [color=blue]
              > - And when I run I expect the program to crash, but it runs fine.[/color]
              [color=blue]
              > $ ./a.out
              > a=1,b=2
              >
              > My questions are:
              > - Is this portable and guranteed by the C standard?
              > - Or is it the manifestation of the C calling convention (the
              > default), hence not portable or
              > implementation dependent.
              > I dug aroung and saw a lot of information regarding 'function pointer
              > compatibility', I guess I
              > don't fully understand that.[/color]

              The conversion is ok, but the call invokes undefined behavior.

              C99 6.3.2.3p8:

              A pointer to a function of one type may be converted to a pointer
              to a function of another type and back again; the result shall
              compare equal to the original pointer. If a converted pointer is
              used to call a function whose type is not compatible with the
              pointed-to type, the behavior is undefined.

              --
              Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
              San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
              We must do something. This is something. Therefore, we must do this.

                Comment

                • MK
                  #5
                  Re: function pointer question

                  Keith,
                  Thanks for the reply. I did repost the correct code in my reply to
                  'raxip' post. I apologize. I see from your post that the behavior is
                  undefined. So what defines the functions 'type'? Is it the just
                  the 'return type' or the whole signature (including the
                  parameters?).
                  The reason I am interested in this, I am looking at a large
                  code-base, and this 'magic' seems to have been used a lot.
                  Thanks.

                    Comment

                    • Ian Collins
                      #6
                      Re: function pointer question

                      MK wrote:[color=blue]
                      >
                      > My questions are:
                      > - Is this portable and guranteed by the C standard?
                      > - Or is it the manifestation of the C calling convention (the
                      > default), hence not portable or
                      > implementation dependent.
                      > I dug aroung and saw a lot of information regarding 'function pointer
                      > compatibility', I guess I
                      > don't fully understand that.
                      >[/color]
                      Based on your second post, I'd say it's undefined behaviour which just
                      happens to wok. Correct me if I'm wrong (I silly thing to say here!),
                      but I don't think there is a standard calling convention.

                      A classic case of an evil cast.

                      --
                      Ian Collins.

                        Comment

                        • Keith Thompson
                          #7
                          Re: function pointer question

                          "MK" <Wavy2Gravy@gma il.com> writes:[color=blue]
                          > Keith,
                          > Thanks for the reply. I did repost the correct code in my reply to
                          > 'raxip' post. I apologize. I see from your post that the behavior is
                          > undefined. So what defines the functions 'type'? Is it the just
                          > the 'return type' or the whole signature (including the
                          > parameters?).
                          > The reason I am interested in this, I am looking at a large
                          > code-base, and this 'magic' seems to have been used a lot.[/color]

                          The type of a function is determined by the return type and
                          parameters.

                          It's possible that your system's calling convention (something not
                          determined by the C language) is such that this kind of thing happens
                          to work. You *might* be better off leaving the code as it is,
                          undefined behavior and all, than fixing it at the risk of introducing
                          new bugs. You'll have to make that judgement yourself.

                          And please read <http://cfaj.freeshell. org/google/>.

                          --
                          Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
                          San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
                          We must do something. This is something. Therefore, we must do this.

                            Comment

                            • Rod Pemberton
                              #8
                              Re: function pointer question


                              "MK" <Wavy2Gravy@gma il.com> wrote in message
                              news:1143085691 .556017.280620@ u72g2000cwu.goo glegroups.com.. .[color=blue]
                              > Keith,
                              > Thanks for the reply. I did repost the correct code in my reply to
                              > 'raxip' post. I apologize. I see from your post that the behavior is
                              > undefined. So what defines the functions 'type'? Is it the just
                              > the 'return type' or the whole signature (including the
                              > parameters?).
                              > The reason I am interested in this, I am looking at a large
                              > code-base, and this 'magic' seems to have been used a lot.
                              > Thanks.
                              >[/color]

                              Those casts make me think of a few things:

                              1) A couple of functions with a different number of arguments were merged
                              together
                              2) The underlying functions are written in another language, i.e., C
                              language with Pascal OS calls

                              The second is common on a number of miniframes.


                              Rod Pemberton


                                Comment

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