为什么每次运行应用程序时都必须使用导出定义 LD_LIBRARY_PATH? [英] Why do I have to define LD_LIBRARY_PATH with an export every time I run my application?

问题描述

我有一些代码使用了一些共享库(gcc 上的 c 代码).编译时，我必须使用 -I 和 -L 显式定义包含和库目录，因为它们不在标准位置.当我尝试运行代码时，出现以下错误:

I have some code that uses some shared libraries (c code on gcc). When compiling I have to explicitly define the include and library directories using -I and -L, since they aren't in the standard places. When I try to run the code, I get the following error:

./sync_test 
./sync_test: error while loading shared libraries: libsync.so: cannot open shared object file: No such file or directory

但是，请执行以下操作，一切正常:

However, do the following, everything works just fine:

export LD_LIBRARY_PATH="/path/to/library/"
./sync_test

现在，奇怪的是，这只起作用一次.如果我再次尝试运行 sync_test，除非我先运行 export 命令，否则我会得到同样的错误.我尝试将以下内容添加到我的 .bashrc 中，但没有任何区别:

Now, the strange part is, this only works once. If I try and run sync_test again I get the same error unless I run the export command first. I tried adding the following to my .bashrc, but it made no difference:

LD_LIBRARY_PATH="/path/to/library/"

推荐答案

使用

export LD_LIBRARY_PATH="/path/to/library/"

否则，在您的 .bashrc 中，它只能用于 bash 而不能用于您启动的任何程序.

in your .bashrc otherwise, it'll only be available to bash and not any programs you start.

在链接时尝试使用 -R/path/to/library/ 标志，它会使程序在该目录中查找，您无需设置任何环境变量.

Try -R/path/to/library/ flag when you're linking, it'll make the program look in that directory and you won't need to set any environment variables.

看起来 -R 仅适用于 Solaris，而您使用的是 Linux.

Looks like -R is Solaris only, and you're on Linux.

另一种方法是将路径添加到 /etc/ld.so.conf 并运行 ldconfig.请注意，这是一项适用于所有动态链接二进制文件的全局更改.

An alternate way would be to add the path to /etc/ld.so.conf and run ldconfig. Note that this is a global change that will apply to all dynamically linked binaries.

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