[Bug middle-end/85599] Function need not be evaluated in logical expression

[sgk at troutmask dot apl.washington.edu]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=85599

--- Comment #15 from Steve Kargl <sgk at troutmask dot apl.washington.edu> ---
On Wed, May 09, 2018 at 10:25:59AM +0000, janus at gcc dot gnu.org wrote:
> 
> --- Comment #14 from janus at gcc dot gnu.org ---
> (In reply to Dominique d'Humieres from comment #10)
> > Am I mistaken to read this as being handled by the middle-end?
> 
> The short-circuiting is finally handled by the middle end, since the front end
> translates .and. into &&. See also comment 6.
> 
> Apparently the standard does neither require nor forbid the short-circuiting
> (see c.l.f. discussion), but I would argue that it would be a more reasonable
> for gfortran to avoid the short-circuiting (by translating to &), at least if
> it is not clear whether the function has side effects.
> 

To be clear.

1) Are you proposing that .AND. should be special-cased to
   force evaluation of both operands?

2) Are you proposing that the operands in rational expressions
   must be evaluated?

3) Are you proposing that the operands for all operators must
   be evaluated?

If the answer to any of the above is 'yes', then add a new
option -fno-short-circuit and implement it to transform

result = op1 binop op2

into

tmp1 = op1
tmp2 = op2
result = tmp1 BINOP tmp2

where BINOP differs from binop in that it knows tmp1 and tmp2
have been evaluated.  To be completely symmetric, you'll need
a similar treatment for unary operators.  That is

result = unop op1

becomes

tmp1 = op1
result = UNOP tmp1 

The flow of the compiler would then be

parse code
if (no short circuit) then
   walk expression trees forcing evaulation of all operands
else
   front optimize pass (walk expression trees)
end if
Give everything to middle-end

Seems like a good method to pessimize performance.