• Brian Gardner (5/5) Oct 01 2006 Does DMC perform floating-point optimizations with the -o option?
• Walter Bright (4/8) Oct 01 2006 To check what kinds it does, it's pretty easy to write up a sample,
• Brian Gardner (15/24) Oct 02 2006 Well, I'm not familiar with assembler code.
• Walter Bright (2/3) Oct 02 2006 (x + x) would be equivalent to (2.00 * x).

Brian Gardner <briangr friberg.us> writes:

Does DMC perform floating-point optimizations with the -o option?

For example: y = 2.0l * (1.0l * x + 0.0l);
after optimization becomes: y = x + x;

Thanks,
Brian

Walter Bright <newshound digitalmars.com> writes:

Brian Gardner wrote:
 Does DMC perform floating-point optimizations with the -o option?
 
 For example: y = 2.0l * (1.0l * x + 0.0l);
 after optimization becomes: y = x + x;

To check what kinds it does, it's pretty easy to write up a sample, 
compile it, then obj2asm the result. For the above case, the two 
statements aren't equivalent.

Brian Gardner <briangr friberg.us> writes:

Walter Bright wrote:
 Brian Gardner wrote:
 Does DMC perform floating-point optimizations with the -o option?

 For example: y = 2.0l * (1.0l * x + 0.0l);
 after optimization becomes: y = x + x;

 
 To check what kinds it does, it's pretty easy to write up a sample, 
 compile it, then obj2asm the result.

Well, I'm not familiar with assembler code.
Anyway I set up a test code:

#include <stdio.h>

int main(int argc, char *argv[]){
  long double y, x = argc;
  y = 1.0l * x + 0.0l;

  return printf("%Lg\n", y);
}

With or without -o, instruction faddp is in the asm code generated from 
obj2asm. So, this means that x + 0.0l is left?
What about for 1.0l * x?

For the above case, the two statements aren't equivalent.

So, 2.0l * x is not equivalent to x + x? Why?

Thanks,
Brian

Walter Bright <newshound digitalmars.com> writes:

Brian Gardner wrote:
 So, 2.0l * x is not equivalent to x + x? Why?

(x + x) would be equivalent to (2.00 * x).