Bootstrapping: don't use convenience symlinks

When `compile.sh` builds bazel using bazel, it
copies the resulting binary to `output/bazel`.
However sometimes the convenience symlink
`bazel-bin` is not created, probably because an
old one is still around and cannot be deleted.

That is clearly a bug, but to work around it, the
bootstrap builder shouldn't attempt to rely on the
creation of these symlinks in the first place.

This change updates compile.sh to use `bazel info`
to locate the `bazel-bin` directory's real path,
and attempt to copy the bazel binary from there.

This works around
https://github.com/bazelbuild/bazel/issues/1827

--
MOS_MIGRATED_REVID=134398451

1 files changed, 7 insertions, 2 deletions

diff --git a/compile.sh b/compile.sh
index 53fc4127b9..484b043a57 100755
--- a/compile.sh
+++ b/compile.sh
@@ -99,8 +99,13 @@ if [ $DO_COMPILE ]; then
   new_step 'Building Bazel with Bazel'
   display "."
   log "Building output/bazel"
-  bazel_build "src:bazel${EXE_EXT}"
-  cp -f "bazel-bin/src/bazel${EXE_EXT}" "output/bazel${EXE_EXT}"
+  bazel_build "src:bazel${EXE_EXT}" \
+    || fail "Could not build Bazel"
+  bazel_bin_path="$(get_bazel_bin_path)/src/bazel${EXE_EXT}"
+  [ -e "$bazel_bin_path" ] \
+    || fail "Could not find freshly built Bazel binary at '$bazel_bin_path'"
+  cp -f "$bazel_bin_path" "output/bazel${EXE_EXT}" \
+    || fail "Could not copy '$bazel_bin_path' to 'output/bazel${EXE_EXT}'"
   chmod 0755 "output/bazel${EXE_EXT}"
   BAZEL="$(pwd)/output/bazel${EXE_EXT}"
 fi