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Re: [PATCH] Fixing improper conversion from sin() to sinf() in optimization mode.

On Fri, 6 Sep 2013, Cong Hou wrote:

> 4: (float) sqrtl ((long double) double_val)  ->  (float) sqrt (double_val)

I don't believe this case is in fact safe even if precision (long double) 
>= precision (double) * 2 + 2 (when your patch would allow it).

The result that precision (double) * 2 + 2 is sufficient for the result of 
rounding the long double value to double to be the same as the result of 
rounding once from infinite precision to double would I think also mean 
the same when rounding of the infinite-precision result to float happens 
once - that is, if instead of (float) sqrt (double_val) you have fsqrt 
(double_val) (fsqrt being the proposed function in draft TS 18661-1 for 
computing a square root of a double value, rounded exactly once to float).  
But here the question isn't whether rounding to long double then float 
gives the same result as rounding to float.  It's whether rounding to long 
double then float gives the same result as rounding to double then float.

Consider, for example, double_val = 0x1.0000020000011p+0.  The square root 
rounded once to float (and so the result if you go via long double and 
long double is sufficiently precise) is 0x1.000002p+0.  But the square 
root rounded to double is 0x1.000001p+0, which on rounding to float 
produces 0x1p+0.

> 5: (double) sqrtl ((long double) float_val)  ->  sqrt ((double) float_val)

(This case, however, *is* safe if long double is sufficiently precise; the 
conversion of float to long double is trivially equivalent to a conversion 
involving an intermediate "double" type, so it reduces to a case for which 
the standard formula involving precisions of just two types applies.)

Joseph S. Myers

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