dojump: Fix up probabilities splitting in dojump.c comparison splitting [PR98212]

dojump: Fix up probabilities splitting in dojump.c comparison splitting [PR98212]

When compiling:
void foo (void);
void bar (float a, float b) { if (__builtin_expect (a != b, 1)) foo (); }
void baz (float a, float b) { if (__builtin_expect (a == b, 1)) foo (); }
void qux (float a, float b) { if (__builtin_expect (a != b, 0)) foo (); }
void corge (float a, float b) { if (__builtin_expect (a == b, 0)) foo (); }
on x86_64, we get (unimportant cruft removed):
bar:    ucomiss %xmm1, %xmm0
        jp      .L4
        je      .L1
.L4:    jmp     foo
.L1:    ret
baz:    ucomiss %xmm1, %xmm0
        jp      .L6
        jne     .L6
        jmp     foo
.L6:    ret
qux:    ucomiss %xmm1, %xmm0
        jp      .L13
        jne     .L13
        ret
.L13:   jmp     foo
corge:  ucomiss %xmm1, %xmm0
        jnp     .L18
.L14:   ret
.L18:   jne     .L14
        jmp     foo
(note for bar and qux that changed with a patch I've posted earlier today).
This is all reasonable, except the last function, the overall jump to
the tail call is predicted unlikely (10%), so it is good jmp foo isn't on
the straight line path, but NaNs are (or should be) considered very unlikely
in the programs, so IMHO the right code (and one emitted with the following
patch) is:
corge:  ucomiss %xmm1, %xmm0
        jp      .L14
        je      .L18
.L14:   ret
.L18:   jmp     foo

Let's discuss the probabilities in the above testcase:
for !and_them it looks all correct, so for
bar we split
if (a != b) goto t; // prob 90%
goto f;
into:
if (a unord b) goto t; // first_prob = prob * cprob = 90% * 1% = 0.9%
if (a ltgt b) goto t; // adjusted prob = (prob - first_prob) / (1 - first_prob) = (90% - 0.9%) / (1 - 0.9%) = 89.909%
and for qux we split
if (a != b) goto t; // prob 10%
goto f;
into:
if (a unord b) goto t; // first_prob = prob * cprob = 10% * 1% = 0.1%
if (a ltgt b) goto t; // adjusted prob = (prob - first_prob) / (1 - first_prob) = (10% - 0.1%) / (1 - 0.1%) = 9.910%
Now, the and_them cases should be probability wise exactly the same
if we swap the f and t labels, because baz
if (a == b) goto t; // prob 90%
goto f;
is equivalent to:
if (a != b) goto f; // prob 10%
goto t;
which is in qux.  This means we could expand baz as:
if (a unord b) goto f; // 0.1%
if (a ltgt b) goto f; // 9.910%
goto t;
But we don't expand it exactly that way, but instead (as the comment says)
as:
if (a ord b) ; else goto f; // first_prob as probability of ;
if (a uneq b) goto t; // adjusted prob
goto f;
So, first_prob.invert () should be 0.1% and adjusted prob should be
1 - 9.910%.
Thus, the right thing is 4 inverts:
prob = prob.invert (); // baz is equivalent to qux with swap(t, f) and thus inverted original prob
first_prob = prob.split (cprob.invert ()).invert ();
// cprob.invert because by doing if (cond) ; else goto f; we effectively invert the condition
// the second invert because first_prob is probability of ; rather than goto f
prob = prob.invert (); // lastly because adjusted prob we want is
// probability of goto t;, while the one from corresponding !and_them case
// would be if (...) goto f; goto t;

2020-12-11  Jakub Jelinek  <jakub@redhat.com>

	PR rtl-optimization/98212
	* dojump.c (do_compare_rtx_and_jump): Change computation of
	first_prob for and_them.  Add comment explaining and_them case.

	* gcc.dg/predict-8.c: Adjust expected probability.