Question

如何在大多数静态模式（linkstatic = 1）中静态链接系统库？我试图使用“-Wl，-Bstatic -lboost_thread -Wl，-Bdynamic”或“-Wl，-Bstatic”，“ - lboost_thread”，“ - WI，-Bdynamic”，但它们都没有奏效。我不想硬编码系统中libboost_thread.a的路径。

cc_binary(
    name = "main",
    srcs = [
        "main.cpp",
    ],
    linkopts = [
        "-lboost_thread",
    ],
)

boost_thread库作为动态库链接。

ldd bazel-bin/main
linux-vdso.so.1
libboost_thread.so.1.54.0 => /usr/lib/x86_64-linux-gnu/libboost_thread.so.1.54.0
libstdc++.so.6 => /usr/lib/x86_64-linux-gnu/libstdc++.so.6
...

Answer 1

在您的WORKSPACE文件中定义外部存储库...

Control

创建boost_thread.BUILD文件

new_local_repository(
    name = "boost_thread",
    path = "/usr/lib/x86_64-linux-gnu",
    build_file = "boost_thread.BUILD"
)

然后在cc_binary规则中添加

cc_library(
   name = "lib",
   srcs = ["libboost_thread.a"],
   visibility = ["//visibility:public"],
)

并投入

deps = ["@boost_thread//:lib",],

安全起见。

Answer 2

基于此问题的答案，Telling gcc directly to link a library statically, "-l:libboost_thread.a" 将静态链接系统库，无需硬编码 libboost_thread.a 在系统中的路径。

cc_binary(
    name = "main",
    srcs = [
        "main.cpp",
    ],
    linkopts = [
        "-l:libboost_thread.a",
        "-l:libboost_system.a",
    ],
)

如何在Bazel中静态链接系统库？

2 个答案: