在 Bash 脚本中调用 C 函数
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Call C function in Bash script
提问by l0b0
Related to question 3451993, is it possible to call a function which is internal to subst.c (in the Bash source code) in a Bash script?
与问题 3451993相关，是否可以在 Bash 脚本中调用 subst.c（在 Bash 源代码中）内部的函数？
采纳答案by Paused until further notice.
Bash supports loadable builtins. You might be able to make use of this to do what you want. See the files in your /usr/share/doc/bash/examples/loadables(or similar) directory.
Bash 支持可加载的内置函数。您也许可以利用它来做您想做的事。查看您/usr/share/doc/bash/examples/loadables（或类似）目录中的文件。
回答by Jay
The simplest way to do this is to write a simple program that collects the input, feeds it to the function, then prints the result. Why don't you tell us what you are attempting to accomplish and perhaps we can suggest an easier way to "skin this cat".
最简单的方法是编写一个简单的程序来收集输入，将其提供给函数，然后打印结果。你为什么不告诉我们你想完成什么，也许我们可以建议一种更简单的方法来“给这只猫剥皮”。
回答by msw
No.
不。
You can't access a function internal to the shell binary from the shell if it is not exported as a shell function.
如果未将其导出为 shell 函数，则无法从 shell 访问 shell 二进制文件内部的函数。
回答by tomlogic
No, you'll have to write a short C program, compile it and call it from the shell.
不，您必须编写一个简短的 C 程序，编译它并从 shell 中调用它。
回答by H_7
This code looks pretty elegant: (from here) Its the same solution @Jay pointed out.
这段代码看起来很优雅：（从这里开始）它与@Jay 指出的解决方案相同。
bash$ cat testing.c
#include <stdio.h>
char* say_hello()
{
return "hello world";
}
float calc_xyzzy()
{
return 6.234;
}
int main(int argc, char** argv)
{
if (argc>1) {
if (argv[1][0] =='1') {
fprintf(stdout,"%s\n",say_hello());
} else if ( argv[1][0] == '2') {
fprintf(stdout,"%g\n",calc_xyzzy());
}
}
return 0;
}
bash$ gcc -o testing testing.c
bash$ ./testing 1
hello world
bash$ ./testing 2
6.234
bash$ var_1="$(./testing 1)"
bash$ var_2="$(./testing 2)"
bash$ echo $var_1
hello world
bash$ echo $var_2
6.234
bash$
