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3118. fpos equality comparison unspecified

Section: [fpos.operations] Status: New Submitter: Jonathan Wakely Opened: 2018-06-04 Last modified: 2022-05-01

Priority: 4

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The fpos requirements do not give any idea what is compared by operator== (even after Daniel's P0759R1 paper). I'd like something to make it clear that return true; is not a valid implementation of operator==(const fpos<T>&, const fpos<T>&). Maybe in the P(o) row state that "p == P(o)" and "p != P(o + 1)", i.e. two fpos objects constructed from the same streamoff values are equal, and two fpos objects constructed from two different streamoff values are not equal.

[2018-06-23 after reflector discussion]

Priority set to 4

[2022-05-01; Daniel comments and provides wording]

The proposed wording does intentionally not use a form involving addition or subtraction to prevent the need for extra wording that ensures that this computed value is well-defined. According to 31.2.2 [stream.types], streamoff is a signed basic integral type, so we know what equality means for such values.

Proposed resolution:

This wording is relative to N4910.

  • Modify in [fpos.operations] as indicated:

    [Drafting note: The return type specification of operator== should be resolved in sync with D2167R2; see also LWG 2114.]

    1. (1.1) — […]

    2. […]

    3. (1.5) — o and o2 refers to a values of type streamoff or const streamoff.

    Table 119: Position type requirements [tab:fpos.operations]
    Expression Return type Operational
    P p(o);
    P p = o;
    Effects: Value-initializes the
    state object.
    Postconditions: p == P(o) is true.
    O(p) streamoff converts to offset P(O(p)) == p
    p == q convertible to bool Remarks: For any two values o and o2, if
    p is obtained from o converted to P or from a copy
    of such P value and if q is obtained from o2
    converted to P or from a copy of such P value, then
    bool(p == q) is true only if o == o2 is true.
    p != q convertible to bool !(p == q)