The solution to 2.84 seems to have a problem

[HITKehaoChen]

I think you should not simply judge ux uy by comparing sx==sy. I think maybe the link is a proper answer.
https://github.com/mofaph/csapp/blob/master/exercise/ex2-83.c

[DreamAndDead]

I think we get the same goal with different expressions.

[zagortenay333]

This solution doesn't handle zeros. x = 000..00 and y = 100..00

[DreamAndDead]

@zagortenay333 did you mean +0 and -0?

This function named float_le, less or equal, so float_le(+0, -0) should return true here.

[zagortenay333]

Yes, but it won't return true when x=+0 and y=-0 since sx != sy and sx > sy is false.

[DreamAndDead]

Yes, you are right. Any elegant way to solve this?

[zagortenay333]

    return
        !(ux<<1 || uy<<1) ||          /* both zero         */
        (sx > sy) ||                  /* x < 0 and y >= 0  */
        (!sx && !sy && ux <= uy) ||   /* x >= 0 and y >= 0 */
        (sx && sx && ux >= uy);       /* x < 0 and y < 0   */

[DreamAndDead]

thanks @LittleCoke @zagortenay333 , how do you think 9d75393

[1036-ce]

Thank you!!

[sci-42ver]

maybe need use

assert(float_le(-0.0, +0.0));
assert(float_le(+0.0, -0.0));

to generate negative zero known from this.

maybe also can use return sx == sy ? (sx == 0 ? ux <= uy : ux >= uy) : sx > sy ? sx > sy : ux+uy==ux-uy && ux+uy==uy-ux; to test which is inspired by above https://github.com/mofaph/csapp/blob/master/exercise/ex2-83.c

[paradox181]

我已收到您的邮件