C++ GCC 不能抱怨未定义的引用吗

发布于04月06日

在什么情况下，GCC not在try 调用组合函数时可能抛出"undefined reference"链接错误消息？

举个例子，有一种情况，这段C代码是由GCC编译链接的:

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

.即使代码中不存在made_up_function_name,anywhere(不是头文件、源文件、声明或任何第三方库).

GCC是否可以在特定条件下接受并编译这种代码，而不涉及实际代码？如果是，哪个？

谢谢

编辑:之前的声明或提到made_up_function_name在其他任何地方都不存在.这意味着整个文件系统的grep -R将only显示确切的单行代码.

推荐答案

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

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

从man ld

--unresolved-symbols=method

确定如何处理未解析的符号.一共有四个方法的可能值:

       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.

共享库本身的行为也可以是

通常，链接器将 for each 报告的未解析符号，但选项--warn-unsolved-symbols可以将此更改为警告.