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  • Re: Regarding assignment to struct

    From Tim Rentsch@tr.17687@z991.linuxsc.com to comp.lang.c on Fri Jun 6 17:44:14 2025
    From Newsgroup: comp.lang.c

    Andrey Tarasevich <noone@noone.net> writes:

    On Tue 5/6/2025 10:36 AM, Waldek Hebisch wrote:

    Keith Thompson <Keith.S.Thompson+u@gmail.com> wrote:

    Andrey Tarasevich <noone@noone.net> writes:
    [...]

    #include <stdio.h>

    struct S { int a[10]; };

    int main()
    {
    struct S a, b = { 0 };
    int *pa, *pb, *pc;

    pa = &a.a[5],
    pb = &b.a[5],
    pc = &(a = b).a[5],
    printf("%p %p %p\n", (void *) pa, (void *) pb, (void *) pc);
    }

    This version has no UB.

    I believe it does. pc points to an element of an object with
    temporary lifetime. The value of pc is then used after the object
    it points to has reached the end of its lifetime. At that point,
    pc has an indeterminate value.

    N3096 6.2.4p2: "If a pointer value is used in an evaluation after
    the object the pointer points to (or just past) reaches the end of
    its lifetime, the behavior is undefined. The representation of a
    pointer object becomes indeterminate when the object the pointer
    points to (or just past) reaches the end of its lifetime."

    Note commas above. Assignment to pc and call to printf are parts
    of a single expression, so use of pc is within lifetime of the
    temporary object.

    Exactly. I thought the nature of the corrections I made (i.e. the
    deliberate usage of comma operator) would be strikingly obvious to the participants of the thread. But alas...

    My own reaction is that the changes were not by themselves
    strikingly obvious. But in combination with the explicit
    statement that "This version has no UB" it seems obvious
    enough.
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