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Re: [ub] Type punning to avoid copying

From: Howard Hinnant <howard.hinnant_at_[hidden]>
Date: Thu, 25 Jul 2013 13:40:36 -0400
On Jul 25, 2013, at 11:45 AM, Gabriel Dos Reis <gdr_at_[hidden]> wrote:

> Howard Hinnant <howard.hinnant_at_[hidden]> writes:
> | It was this SO question that started this thread:
> |
> | http://stackoverflow.com/q/17789928/576911
> OK, this version obviously breaks the rule and invoked undefined behavior.
> We won't go over that again. But, let me point out that this particular
> type-based alias analysis in GCC has a long history and tradition -- and
> most certainly longer in high-performance compilers. When it was
> implemented in GCC by Mark Mitchell (I think), we found that type-based
> alias analysis found lot of broken (C++) codes. You can search GCC
> mailing list archive for the long discussions, and also the web. So
> it was moved under a dedicated -fstrict-aliasing flag, then after
> a while activated at -O2. This is ancient history, by which I mean it
> happened *15* years ago at least, circa 1998. Independently, people wanted a
> dedicated flag for activating the warning; that was introduced more than
> a decade ago, in 2002.
> | I'm curious: The accepted answer uses memcpy and the claim is that
> | this is a correct answer to the question. That is it does not exhibit
> | undefined behavior. My current understanding is that I agree with
> | this answer. But I wanted to check here. Do people here agree that:
> |
> | http://stackoverflow.com/a/17790026/576911
> |
> | does not break the aliasing rules, or otherwise invoke undefined behavior?
> This version does not break aliasing rules. The question of whether it
> invokes undefined behavior is less obvious. The current rules guarantee
> that if you memcpy() to a place, and memcpy() the same bytes back, you
> get the original value and that is well defined. The rules also say
> that if you don't initialize an object, but you do a read access you
> invoke undefined behavior unless the object is of character type or array
> thereof. Now, there is an interesting twist here, which is that int32_t
> admits no trap representation (just like characters), I cannot find a
> rule a rule that says that even if int32_t object isn't initialized it
> is OK, or whether memcpy() into the object is initialization. Note, the
> first memcpy() isn't from an int32_t.
> If that analysis is correct, I would welcome a rule in the language that
> says that memcpy() into an object of type that admits no trap
> representation is equivalent to initialization with value obtained from
> appropriate interpretation of the value representation bits of the
> "initializing byte" sequence. I think this won't hinder existing
> program transformation since memcpy() is essentially understood as
> ending previous object lifetime and starting a new one.

Ok, thanks. I'm copying the entire thing here for reference:

float InverseSquareRoot2(float x)
    float xhalf = 0.5f*x;
    int32_t i;
    assert(sizeof(x) == sizeof(i));
    std::memcpy(&i, &x, sizeof(i));
    i = 0x5f3759df - (i>>1);
    std::memcpy(&x, &i, sizeof(i));
    x = x*(1.5f - xhalf*x*x);
    return x;

If we did that (make InverseSquareRoot2 implementation defined, well defined, whatever), would that also confer the same status to this rewrite of InverseSquareRoot2?

float InverseSquareRoot4(float x)
    union U
        float as_float;
        int32_t as_int;
    U u = {x};
    U tmp;
    float xhalf = 0.5f*x;
    std::memcpy(&tmp.as_int, &u.as_float, sizeof(U));
    std::memcpy(&u.as_int, &tmp.as_int, sizeof(U));
    u.as_int = 0x5f3759df - (u.as_int>>1);
    std::memcpy(&tmp.as_float, &u.as_int, sizeof(U));
    std::memcpy(&u.as_float, &tmp.as_float, sizeof(U));
    u.as_float = u.as_float*(1.5f - xhalf*u.as_float*u.as_float);
    return u.as_float;


Received on 2013-07-25 19:40:32