GCC可以不抱怨未定义的引用吗？

Question

在什么情况下,GCC 在尝试调用伪造函数时不会抛出"未定义的引用"链接错误消息？

例如,GCC编译和链接此C代码的情况:

void function()
{
    made_up_function_name();
    return;
}

...即使代码中made_up_function_name没有任何地方(不是标题,源文件,声明或任何第三方库).

GCC是否可以在某些条件下接受和编译这种代码,而无需触及实际代码？如果是这样,哪个？

谢谢.

编辑:之前没有任何声明或提及made_up_function_name.这意味着grep -R整个文件系统中的一个只显示完整的单行代码.

Answer 1

是的,可以使用--unresolved-symbols链接器选项来避免报告未定义的引用.

g++ mm.cpp -Wl,--unresolved-symbols=ignore-in-object-files

从 man ld

--unresolved符号=方法

确定如何处理未解析的符号.方法有四种可能的值:

       ignore-all
           Do not report any unresolved symbols.

       report-all
           Report all unresolved symbols.  This is the default.

       ignore-in-object-files
           Report unresolved symbols that are contained in shared
           libraries, but ignore them if they come from regular object
           files.

       ignore-in-shared-libs
           Report unresolved symbols that come from regular object
           files, but ignore them if they come from shared libraries.  This
           can be useful when creating a dynamic binary and it is known
           that all the shared libraries that it should be referencing
           are included on the linker's command line.

共享库的行为也可以通过 - [no-] allow-shlib-undefined选项控制.

通常,链接器将为每个报告的未解析符号生成错误消息,但选项--warn-unresolved-symbols可以将此更改为警告.