Revert "ifcvt: replace C++ sort with vec::qsort [PR109154]"

This reverts commit 19610580d49f3d2d8d511fba55b39efa0764dfc2.

It broke bootstrap because of a recently added assert, and
fixing will take time.

1 files changed, 6 insertions, 23 deletions

diff --git a/gcc/tree-if-conv.cc b/gcc/tree-if-conv.cc
index 281984d..a8c9159 100644
--- a/gcc/tree-if-conv.cc
+++ b/gcc/tree-if-conv.cc
@@ -80,6 +80,7 @@ along with GCC; see the file COPYING3.  If not see
      <L18>:;
 */
 
+#define INCLUDE_ALGORITHM
 #include "config.h"
 #include "system.h"
 #include "coretypes.h"
@@ -2044,28 +2045,6 @@ gen_phi_nest_statement (gphi *phi, gimple_stmt_iterator *gsi,
   return lhs;
 }
 
-typedef std::pair <tree, std::pair <unsigned, unsigned>> ArgEntry;
-static int
-cmp_arg_entry (const void *p1, const void *p2)
-{
-  const ArgEntry sval1 = *(const ArgEntry *)p1;
-  const ArgEntry sval2 = *(const ArgEntry *)p2;
-  auto x1 = sval1.second;
-  auto x2 = sval2.second;
-
-  if (x1.first < x2.first)
-    return -1;
-  else if (x1.first > x2.first)
-    return 1;
-
-  if (x1.second < x2.second)
-    return -1;
-  else if (x1.second > x2.second)
-    return 1;
-
-  return 0;
-}
-
 /* Replace a scalar PHI node with a COND_EXPR using COND as condition.
    This routine can handle PHI nodes with more than two arguments.
 
@@ -2207,6 +2186,7 @@ predicate_scalar_phi (gphi *phi, gimple_stmt_iterator *gsi)
   /* Determine element with max number of occurrences and complexity.  Looking at only
      number of occurrences as a measure for complexity isn't enough as all usages can
      be unique but the comparisons to reach the PHI node differ per branch.  */
+  typedef std::pair <tree, std::pair <unsigned, unsigned>> ArgEntry;
   auto_vec<ArgEntry> argsKV;
   for (i = 0; i < args.length (); i++)
     {
@@ -2224,7 +2204,10 @@ predicate_scalar_phi (gphi *phi, gimple_stmt_iterator *gsi)
     }
 
   /* Sort elements based on rankings ARGS.  */
-  argsKV.qsort (cmp_arg_entry);
+  std::sort(argsKV.begin(), argsKV.end(), [](const ArgEntry &left,
+					     const ArgEntry &right) {
+    return left.second < right.second;
+  });
 
   for (i = 0; i < args.length (); i++)
     args[i] = argsKV[i].first;