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[PATCH] Fix SPARC movdf ICE
- To: gcc-patches at gcc dot gnu dot org
- Subject: [PATCH] Fix SPARC movdf ICE
- From: Jakub Jelinek <jakub at redhat dot com>
- Date: Fri, 9 Nov 2001 15:34:07 +0100
- Reply-To: Jakub Jelinek <jakub at redhat dot com>
Hi!
The following testcase ICEs on SPARC at -O2. The problem is that SPARC
movdf_* insns allow '*r' <- '*rFo' constraints, but when reload does
anything with it and in turn calls gen_movdf, we ICE because it cannot be
validized any more, although sparc.md has splitters which will build the
constant in general registers and won't need the DF constant in memory.
Commited.
2001-11-09 Jakub Jelinek <jakub@redhat.com>
* config/sparc/sparc.md (movdf): Avoid calling validize_mem during
or after reload.
* gcc.c-torture/compile/20011109-1.c: New test.
--- gcc/config/sparc/sparc.md.jj Wed Oct 24 21:25:07 2001
+++ gcc/config/sparc/sparc.md Fri Nov 9 15:28:04 2001
@@ -3255,6 +3255,11 @@
&& fp_zero_operand (operands[1], DFmode))
goto movdf_is_ok;
+ /* We are able to build any DF constant in integer registers. */
+ if (REGNO (operands[0]) < 32
+ && (reload_completed || reload_in_progress))
+ goto movdf_is_ok;
+
operands[1] = validize_mem (force_const_mem (GET_MODE (operands[0]),
operands[1]));
}
--- gcc/testsuite/gcc.c-torture/compile/20011109-1.c.jj Fri Nov 9 15:15:34 2001
+++ gcc/testsuite/gcc.c-torture/compile/20011109-1.c Fri Nov 9 15:12:42 2001
@@ -0,0 +1,51 @@
+typedef struct { short x[4]; } S;
+typedef struct { unsigned int a, b, c; S *d; } T;
+
+S *(*foo) (T *, int, int, int, int);
+unsigned short *(*bar)(const T *);
+unsigned short baz(T *,const int);
+
+T *die (void)
+{
+ typedef struct { unsigned int a, b, e; double f, g; } U;
+
+ char h[8], i[2053], j[2053];
+ double k, l, m;
+ U n;
+ T *o;
+ unsigned short p;
+ int q, r;
+ long s;
+ unsigned short *t;
+ S *u;
+ unsigned char *v, *w;
+ unsigned int x;
+
+ o = 0;
+ for (x = 0; x < n.e; x++)
+ {
+ l = 1.0;
+ if (n.g - n.f <= 1.0)
+ l = ((1 << o->c) - 1) / (n.g - n.f);
+ v = w;
+ for (r = o->b - 1; r >= 0; r--)
+ {
+ u = foo (o, 0, r, o->a, 1);
+ if (!u)
+ break;
+ t = bar (o);
+ for (q = 0; q < (int) o->a; q++)
+ {
+ h[0] = *v;
+ s = *v++;
+ k = (double) s;
+ m = l*k;
+ p = m < 0 ? 0 : m > (1 << o->c) - 1 ? (1 << o->c) - 1 : m + 0.5;
+ p = baz (o,p);
+ t[q] = p;
+ *u++ = o->d[p];
+ }
+ }
+ }
+ return o;
+}
Jakub