bazel - Genrule 不会在最终输出文件中输出“$@”符号

Question

我正在写一个如下的类：

genrule(
name = "create_run_script",
outs = ["run_script.sh"],
executable = True,
cmd = """
      cat > $@ << EOF
      #!/bin/bash
      cd ../dir/
      exec ./program "$$@"
      EOF
      """,

)

当我执行此操作时，我看到了输出

    # @external_repo//:create_run_script [action 'Executing genrule @ external_repo//:create_run_script']
(cd /home/marc/.cache/bazel/_bazel_marc/0877f3fef7c185b84693d3a53e00a8be/execroot/zoox && \
  exec env - \
    FLAGS_minloglevel=1 \
    LD_LIBRARY_PATH='' \
    PATH=/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin \
  /bin/bash -c 'source external/bazel_tools/tools/genrule/genrule-setup.sh; 
          cat > bazel-out/k8-fastbuild/bin/external/external_repo/run_script.sh << EOF
          #!/bin/bash
          cd ../external_repo/
          exec ./program "$@"
          EOF
          ')

但是，当我打开实际的 shell 脚本文件时，我看到

      #!/bin/bash
      cd ../external_repo/
      exec ./program ""
      EOF
      

$@ 符号不再存在！如何使用 Bazel genrule 指令创建一个可以接收来自 shell 的输入的 shell 脚本？

Answer 1

创建该脚本的 shell 调用正在扩展$@；heredoc 需要禁用扩展：

genrule(
name = "create_run_script",
outs = ["run_script.sh"],
executable = True,
cmd = """
      cat > $@ << 'EOF'
      #!/bin/bash
      cd ../dir/
      exec ./program "$$@"
EOF
      """,
)

（注意周围的引号'EOF'。）