如何在Bazel中获取在execpath中创建的文件的文件对象？

def _generate_file(ctx):
    # 通过运行可执行文件生成文件
    ctx.actions.run(
        executable = "//path/to/executable",
        outputs = ["file.txt"],
        arguments = [],
    )

def generate_rule():
    native.genrule(
        name = "generate_file",
        srcs = [],
        outs = ["file.txt"],
        cmd = "_generate_file $@",
    )

def use_file_rule():
    # 获取生成的文件对象
    file_obj = ctx.actions.declare_file("my_file", output_ = "file.txt")
    # 使用文件对象进行进一步操作
    # ...

def rules():
    native.bind(
        name = "use_file",
        implementation = use_file_rule,
        attrs = {},
    )