编译没有的libc [英] Compiling without libc

问题描述

我想编译我的C code无（G）的libc。我怎样才能将其停用，哪些功能取决于它？

I want to compile my C-code without the (g)libc. How can I deactivate it and which functions depend on it?

我试过-nostdlib但它没有帮助：code是编译和运行，但我仍然可以找到自己的可执行文件的hexdump都libc中的名称

I tried -nostdlib but it doesn't help: The code is compilable and runs, but I can still find the name of the libc in the hexdump of my executable.

推荐答案

如果您编译code。与-nostdlib，你将无法调用任何C库函数（当然），但你也不要得到的，常规的C引导code。特别是，在Linux的节目的实际进入点是不main（）中，而是一个函数调用_start（）。该标准库通常提供一个版本，这个运行一些初始化code，然后调用的main（）。

If you compile your code with -nostdlib, you won't be able to call any C library functions (of course), but you also don't get the regular C bootstrap code. In particular, the real entry point of a program on linux is not main(), but rather a function called _start(). The standard libraries normally provide a version of this that runs some initialization code, then calls main().

尝试用gcc编译-nostdlib这样的：

Try compiling this with gcc -nostdlib:

void _start() {

    /* main body of program: call main(), etc */

    /* exit system call */
    asm("movl $1,%eax;"
        "xorl %ebx,%ebx;"
        "int  $0x80"
    );
}

在_start（）函数应始终以呼叫结束退出（或其他非返回系统调用，如执行）。上面的例子直接与内联汇编调用系统调用，因为通常的出口（）不可用。

The _start() function should always end with a call to exit (or other non-returning system call such as exec). The above example invokes the system call directly with inline assembly since the usual exit() is not available.

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