首页 > 解决方案 > 将环境变量从 makefile 导出到 shell

问题描述

我想修改和导出“LD_LIBRARY_PATH”环境变量,以便将 libjvm.so 与我的代码链接起来。下面是我的Makefile:

 all: run

 helloWorld.class: helloWorld.java
     javac helloWorld.java

 hello_world: hello_world.c
     gcc -L/usr/lib/jvm/java-8-openjdk-amd64/jre/lib/amd64/server/ -I/usr/lib/jvm/java-8-openjdk-amd64/include -I/usr/lib/jvm/java-8-openjdk-amd64/include/linux/ hello_world.c -o hello_world_c_exec -ljvm

 run: helloWorld.class hello_world 
     export LD_LIBRARY_PATH="/usr/lib/jvm/java-8-openjdk-amd64/jre/lib/amd64/server/"
     ./hello_world_c_exec

 clean:
     rm -f helloWorld.class hello_world

在名为“run”的目标中,我导出“LD_LIBRARY_PATH”环境变量。但它不会导出到我当前的外壳。

我收到以下错误:./hello_world_c_exec:加载共享库时出错:libjvm.so:无法打开共享对象文件:没有这样的文件或目录

如何使用 makefile 导出“LD_LIBRARY_PATH”?

标签: makefilegnu-make

解决方案


配方的每一行通常在单独的 shell 中执行,这意味着您的export行进入与您的./hello_world_c_exec行不同的 shell。你自己看:

$ cat Makefile
JVM_DIR := /tools/opt/jdk-10.0.2/lib/server

foo: LDLIBS := -ljvm
foo: LDFLAGS := -L$(JVM_DIR)

.PHONY: run
run: foo
        export LD_LIBRARY_PATH=$(JVM_DIR)
        $(<D)/$(<F)

输出:

$ make run -d
...
Must remake target 'run'.
export LD_LIBRARY_PATH=/tools/opt/jdk-10.0.2/lib/server
Putting child 0x81ee20 (run) PID 18709 on the chain.
Live child 0x81ee20 (run) PID 18709
Reaping winning child 0x81ee20 PID 18709
./foo
Live child 0x81ee20 (run) PID 18710
./foo: error while loading shared libraries: libjvm.so: cannot open shared object file: No such file or directory
Reaping losing child 0x81ee20 PID 18710
make: *** [Makefile:9: run] Error 127

请注意,为每个单独的行启动了两个单独的进程(18709 和 18710)。如果它在单个 shell 中运行,您可以通过以下几种方式使其工作:

  1. 使其成为单个脚本:
$ cat Makefile
JVM_DIR := /tools/opt/jdk-10.0.2/lib/server

foo: LDLIBS := -ljvm
foo: LDFLAGS := -L$(JVM_DIR)

.PHONY: run
run: foo
        export LD_LIBRARY_PATH=$(JVM_DIR); \
        $(<D)/$(<F)

$ make run -d
...
Must remake target 'run'.
export LD_LIBRARY_PATH=/tools/opt/jdk-10.0.2/lib/server; \
./foo
Putting child 0x21aa780 (run) PID 22009 on the chain.
Live child 0x21aa780 (run) PID 22009
Reaping winning child 0x21aa780 PID 22009
Removing child 0x21aa780 PID 22009 from chain.
Successfully remade target file 'run'.
  1. 直接在命令行中设置:
$ cat Makefile
JVM_DIR := /tools/opt/jdk-10.0.2/lib/server

foo: LDLIBS := -ljvm
foo: LDFLAGS := -L$(JVM_DIR)

.PHONY: run
run: foo
        LD_LIBRARY_PATH=$(JVM_DIR) $(<D)/$(<F)

$ make run
LD_LIBRARY_PATH=/tools/opt/jdk-10.0.2/lib/server ./foo
  1. 打开One Shell以便配方在单个 shell 中执行,而不是每行单独的 shell:
$ cat Makefile
JVM_DIR := /tools/opt/jdk-10.0.2/lib/server

foo: LDLIBS := -ljvm
foo: LDFLAGS := -L$(JVM_DIR)

.ONESHELL:

.PHONY: run
run: foo
        export LD_LIBRARY_PATH=$(JVM_DIR)
        $(<D)/$(<F)

$ make run -d
...
Must remake target 'run'.
export LD_LIBRARY_PATH=/tools/opt/jdk-10.0.2/lib/server
./foo
Putting child 0x25258f0 (run) PID 31367 on the chain.
Live child 0x25258f0 (run) PID 31367
Reaping winning child 0x25258f0 PID 31367
Removing child 0x25258f0 PID 31367 from chain.
Successfully remade target file 'run'.

就个人而言,我会选择不同的方法。

首先,设置LD_LIBRARY_PATH很麻烦,并且取决于各种设置可能会导致其他事情失败,例如当您的系统已经设置LD_LIBRARY_PATH为其他内容时;至少您应该附加到变量而不是将其设置为严格的值。

其次,这仍然存在在您的Makefile. 即使构建了二进制文件,仅通过简单的调用来运行它也不是那么简单:

$ make run
cc   -L/tools/opt/jdk-10.0.2/lib/server  foo.c  -ljvm -o foo
export LD_LIBRARY_PATH=/tools/opt/jdk-10.0.2/lib/server
./foo
$ ./foo
./foo: error while loading shared libraries: libjvm.so: cannot open shared object file: No such file or directory

相反,我会告诉链接器将提示嵌入到哪里可以找到库,这样LD_LIBRARY_PATH就根本不需要了:

$ cat Makefile
JVM_DIR := /tools/opt/jdk-10.0.2/lib/server

foo: LDLIBS := -ljvm
foo: LDFLAGS := -L$(JVM_DIR) -Wl,-rpath,$(JVM_DIR)

.PHONY: run
run: foo
        $(<D)/$(<F)

$ make run
cc   -L/tools/opt/jdk-10.0.2/lib/server -Wl,-rpath,/tools/opt/jdk-10.0.2/lib/server  foo.c  -ljvm -o foo
./foo

$ ./foo
$ echo $?
0

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