Re: Why is using single-precision slower than using double-precision

Did you used float constant when you were calculating in single precision?
Remenber that floating points constants are double and that if you have
float x,y;
y = 5.0 * x; here you have (double) * (float) -> (float) so in C you have the
following:
y = (float)(5.0 * (double)x) so you have two conversions..


On the other hand if you have
float x,y;
y = 5.0f * x;
              ^a single-precision constant
  Here you have an operation between two float, I thought that the
compiler would generate the following (float) * (float) -> (float). But everyone
here is telling that the compiler will generate
y = (float)((double)5.0f * (double)x)
^^^of course this isn't a true conversion here.


Well if you want to know what's going on, the only way is to look at the assembly


I've compiled on a sun y = 5.0f * x; (y,x float) with gcc -O2
  the multiply instruction is
fmuls %f0,%f2,%f0 (single precision multiply) and it seems that there
isn't any conversion (I don't know sparc assembly...).


So why all these people talked about an automatic conversion ?
Because they are too lazy to insert a "f" after their constants :-).
But it may depends on the compiler though.


Regards.


Renaud HEBERT
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