Restricted pointer parameters in printf()

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  • Peter Ammon

    Restricted pointer parameters in printf()

    It's my understanding that the printf() function is declared as

    int printf(const char * restrict format, ...);

    in stdio.h. And loosely speaking, if a parameter is declared as
    restricted, then accesses to the object must go through that parameter.

    Does this mean that

    printf("%s", "%s");

    is illegal if the string literals are coalesced into the same pointer?
    printf() would be accessing the same object through different
    parameters, one of which is declared as restricted.

    Another possibility is

    printf("%s is tricky", "is tricky");

    since the string literal objects can again overlap even though their
    pointers have different values. Is that code potentially illegal?

    Thanks for your thoughts,
    -Peter

    --
    Pull out a splinter to reply.
  • Jack Klein

    #2
    Re: Restricted pointer parameters in printf()

    On Sat, 10 Apr 2004 04:46:37 GMT, Peter Ammon
    <gershwin@splin termac.com> wrote in comp.lang.c:
    [color=blue]
    > It's my understanding that the printf() function is declared as
    >
    > int printf(const char * restrict format, ...);
    >
    > in stdio.h. And loosely speaking, if a parameter is declared as
    > restricted, then accesses to the object must go through that parameter.
    >
    > Does this mean that
    >
    > printf("%s", "%s");
    >
    > is illegal if the string literals are coalesced into the same pointer?
    > printf() would be accessing the same object through different
    > parameters, one of which is declared as restricted.
    >
    > Another possibility is
    >
    > printf("%s is tricky", "is tricky");
    >
    > since the string literal objects can again overlap even though their
    > pointers have different values. Is that code potentially illegal?
    >
    > Thanks for your thoughts,
    > -Peter[/color]

    With the exception of the destination strings for (v)s(n)printf() ,
    members of the *printf() family never modify memory passed in via a
    pointer to char.

    The restrict keyword is completely unnecessary and has no effect on
    objects that are not modified. It's purpose is to allow the compiler
    to perform optimizations based on the fact that modifying one object
    is guaranteed not to change the value of some other object. There is
    no possibility of that happening when the objects are not modified.

    --
    Jack Klein
    Home: http://JK-Technology.Com
    FAQs for
    comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
    comp.lang.c++ http://www.parashift.com/c++-faq-lite/
    alt.comp.lang.l earn.c-c++

    Comment

    • Peter Ammon

      #3
      Re: Restricted pointer parameters in printf()

      Jack Klein wrote:[color=blue]
      > On Sat, 10 Apr 2004 04:46:37 GMT, Peter Ammon
      > <gershwin@splin termac.com> wrote in comp.lang.c:
      >
      >[color=green]
      >>It's my understanding that the printf() function is declared as
      >>
      >>int printf(const char * restrict format, ...);
      >>
      >>in stdio.h. And loosely speaking, if a parameter is declared as
      >>restricted, then accesses to the object must go through that parameter.
      >>
      >>Does this mean that
      >>
      >>printf("%s" , "%s");
      >>
      >>is illegal if the string literals are coalesced into the same pointer?
      >>printf() would be accessing the same object through different
      >>parameters, one of which is declared as restricted.
      >>
      >>Another possibility is
      >>
      >>printf("%s is tricky", "is tricky");
      >>
      >>since the string literal objects can again overlap even though their
      >>pointers have different values. Is that code potentially illegal?
      >>
      >>Thanks for your thoughts,
      >>-Peter[/color]
      >
      >
      > With the exception of the destination strings for (v)s(n)printf() ,
      > members of the *printf() family never modify memory passed in via a
      > pointer to char.
      >
      > The restrict keyword is completely unnecessary and has no effect on
      > objects that are not modified.[/color]

      Why's that?
      [color=blue]
      > It's purpose is to allow the compiler
      > to perform optimizations based on the fact that modifying one object
      > is guaranteed not to change the value of some other object. There is
      > no possibility of that happening when the objects are not modified.[/color]

      Correct me if I'm wrong, but I thought that the restrict keyword was
      defined in terms of "accesses," which would includes reads as well as
      writes?

      --
      Pull out a splinter to reply.

      Comment

      • Jack Klein

        #4
        Re: Restricted pointer parameters in printf()

        On Sat, 10 Apr 2004 06:50:48 GMT, Peter Ammon
        <gershwin@splin termac.com> wrote in comp.lang.c:
        [color=blue]
        > Jack Klein wrote:[color=green]
        > > On Sat, 10 Apr 2004 04:46:37 GMT, Peter Ammon
        > > <gershwin@splin termac.com> wrote in comp.lang.c:
        > >
        > >[color=darkred]
        > >>It's my understanding that the printf() function is declared as
        > >>
        > >>int printf(const char * restrict format, ...);
        > >>
        > >>in stdio.h. And loosely speaking, if a parameter is declared as
        > >>restricted, then accesses to the object must go through that parameter.
        > >>
        > >>Does this mean that
        > >>
        > >>printf("%s" , "%s");
        > >>
        > >>is illegal if the string literals are coalesced into the same pointer?
        > >>printf() would be accessing the same object through different
        > >>parameters, one of which is declared as restricted.
        > >>
        > >>Another possibility is
        > >>
        > >>printf("%s is tricky", "is tricky");
        > >>
        > >>since the string literal objects can again overlap even though their
        > >>pointers have different values. Is that code potentially illegal?
        > >>
        > >>Thanks for your thoughts,
        > >>-Peter[/color]
        > >
        > >
        > > With the exception of the destination strings for (v)s(n)printf() ,
        > > members of the *printf() family never modify memory passed in via a
        > > pointer to char.
        > >
        > > The restrict keyword is completely unnecessary and has no effect on
        > > objects that are not modified.[/color]
        >
        > Why's that?
        >[color=green]
        > > It's purpose is to allow the compiler
        > > to perform optimizations based on the fact that modifying one object
        > > is guaranteed not to change the value of some other object. There is
        > > no possibility of that happening when the objects are not modified.[/color]
        >
        > Correct me if I'm wrong, but I thought that the restrict keyword was
        > defined in terms of "accesses," which would includes reads as well as
        > writes?[/color]

        The restrict keyword has its roots in the proposed noalias keyword,
        which was proposed, disagreed over vehemently, and finally not
        included in the original C standard ANSI89/ISO 90.

        A weakness of C compared to other languages is the possibility of
        aliasing. Let's take a trivial program like this:

        #include <stdio.h>

        int arr1 [5] = { 2, 4, 6, 8, 10 };
        int arr2 [5] = { 2, 4, 6, 8, 10 };

        void some_func(int *a, int *b, int count)
        {
        while (count--)
        {
        if (*a == *b)
        {
        *a = *b + 2;
        }
        ++a;
        }
        }

        int main(void)
        {
        int b = 2;

        some_func(arr1, &b, 5);
        some_func(arr2, arr2, 5);
        printf("arr1 = { %d, %d, %d, %d, %d }\n",
        arr1[0],arr1[1],arr1[2],arr1[3],arr1[4]);
        printf("arr2 = { %d, %d, %d, %d, %d }\n",
        arr2[0],arr2[1],arr2[2],arr2[3],arr2[4]);
        return 0;
        }

        A simple optimization in some_func() would be for the compiler to read
        *b once, and cache it in a local variable or register, instead of
        reading it each time. But what if b is the address of one of the ints
        in the range a[0] to a[count - 1]? Let's run it and look at the
        output:

        arr1 = { 4, 4, 6, 8, 10 }
        arr2 = { 4, 6, 6, 8, 10 }

        Both calls to some_func() were made with a count of 5 and an input
        array of 5 ints containing 2, 4, 6, 8, and 10. And finally, both
        calls were made with a pointer to an int having the value 2.

        If the compiler performed the optimization of reading *b only once and
        cacheing the value, both arrays would have had identical values after
        being processed by some_func(), which would have been wrong. When
        some_func() is called the second time, b points to arr2[0], so the
        first pass of the loop causes *b to be modified. This causes *b to
        match *a on the second pass of the loop, and the code correctly
        modifies both arr2[0] and arr2[1].

        The price of this flexibility in C is the fact that the compiler
        cannot perform the optimization of caching the value of b, since
        modifying an object through pointer a could modify what b points to.

        If b was defined as a pointer to a const int, the compiler could
        perform that optimization. On the other hand, if b was restrict
        qualified, the compiler could perform that optimization. If it did,
        the results of the second function call would be wrong, but that would
        be the programmer's fault, for invoking undefined behavior.

        The description of the restrict keyword in the standard is rather
        complex, but it still boils down to informing the compiler that the
        value of an object will not be modified unexpectedly via a different
        lvalue. This allows, but does not require, the compiler to perform
        certain optimizations when the programmer specifies, via the restrict
        keyword, that an object will not be modified through another pointer.

        One of the examples from the standard might be helpful in
        understanding the intent:

        ====
        10 EXAMPLE 3 The function parameter declarations
        void h(int n, int * restrict p, int * restrict q, int * restrict r)
        {
        int i;
        for (i = 0; i < n; i++)
        p[i] = q[i] + r[i];
        }
        illustrate how an unmodified object can be aliased through two
        restricted pointers. In particular, if a and b are disjoint arrays, a
        call of the form h(100, a, b, b) has defined behavior, because array b
        is not modified within function h.
        ====

        So even though the term "access" is used several times in the
        definition of the restrict qualifier, it has no meaning for values
        that are not modified at all within the scope of the qualifier.

        --
        Jack Klein
        Home: http://JK-Technology.Com
        FAQs for
        comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
        comp.lang.c++ http://www.parashift.com/c++-faq-lite/
        alt.comp.lang.l earn.c-c++

        Comment

        • Chris Torek

          #5
          Re: Restricted pointer parameters in printf()

          In article <news:N_Kdc.501 07$xc6.38503@ne wssvr25.news.pr odigy.com>
          Peter Ammon <gershwin@splin termac.com> writes:[color=blue]
          >It's my understanding that the printf() function is declared as
          >
          >int printf(const char * restrict format, ...);
          >
          >in stdio.h. And loosely speaking, if a parameter is declared as
          >restricted, then accesses to the object must go through that parameter.[/color]

          Yes, but this is "too loosely". :-)
          [color=blue]
          >Does this mean that
          > printf("%s", "%s");
          >is illegal if the string literals are coalesced into the same pointer?
          >... Another possibility is
          > printf("%s is tricky", "is tricky");[/color]

          Both of these calls are OK, as I think Jack Klein noted elsethread.

          If you want a simpler, yet still loose but not (I hope) *too* loose,
          way to describe "restrict", think of it as you saying to the
          compiler:

          OK, here is this pointer "p", that has the "restrict" on it. I
          promise you, Mr Compiler, that I will not change the object at
          p[i] without actually *writing* p[i], so that you can make
          optimization assumptions.

          In other words, the compiler is allowed to assume that any change
          to *(p + i) for any integer i happens by writing to *(p + j) for
          some integer j such that j==i. (The draft has additional wording
          that lets the programmer use expressions like ((p+2)[i-2]) as well,
          for instance, but this captures the essence.)

          If you never write on format[i] at all -- as is the case here,
          since format[i] is const-qualified -- the compiler can assume that
          format[i] never changes while printf() is doing its thing internally.

          Optimization of variable accesses is mostly a matter of clever
          cacheing, including "reading ahead" (so that inputs are available
          by the time they are needed) and "writing behind" (so that outputs
          do not cause the program to stop and wait for the output to finish).
          If the only kind of cacheing you ever do is read-related, everything
          "just works" -- your caches could even wind up with multiple copies
          of a single variable, but they all have the same value anyway, so
          who cares? :-)

          Unfortunately, there are some subtler interactions when one includes
          write steps. Consider the (not very good) routine:

          #define OPERATE ... /* some operation here */
          void operate(double *result, double *inputs, size_t n) {

          *result = 0.0;
          for (size_t i = 0; i < n; i++)
          *result = *result OPERATE inputs[i];
          }

          followed by a call of the form:

          double buf[N]; /* N >= 11 */
          operate(buf + 10, buf, 11);

          This tries to put the result in buf[10] while using buf[10] as one
          of the inputs.

          With no "restrict" qualifiers, this code is OK -- when i reaches 10
          inside operate(), buf[10] is the accumulated result-so-far, and
          it is fed back into the operation.

          If you were to write "double *restrict result", the differences
          between various kinds of caching begin to show up: if the input[]
          array is read too soon, or the output *result pointer is not written
          soon enough, input[10] will not hold the (presumably) desired value
          at the right time.

          Since I do not have the final C99 standard, I am not willing to
          say what it might require in these cases. I *am* sure, however,
          that for read-only "restrict"e d pointers (as in the printf()
          example), it *is* OK to have aliases going on (despite the wording
          in the draft I use, which seems to forbid it).
          --
          In-Real-Life: Chris Torek, Wind River Systems
          Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) +1 801 277 2603
          email: forget about it http://web.torek.net/torek/index.html
          Reading email is like searching for food in the garbage, thanks to spammers.

          Comment

          • Peter Ammon

            #6
            Re: Restricted pointer parameters in printf()

            Thanks to both Jack Klein and Chris Torek for clarifying the issues. I
            also lack the official C standard and was confused by the draft. Time
            to "upgrade" I suppose. :)


            --
            Pull out a splinter to reply.

            Comment

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