Question

我正在学习c ++，正在尝试在linux上加载共享库（.so）。

运行以下代码时出现分段错误。

当我尝试使用valgrind运行控制台应用程序时，我得到以下内容：

valgrind ./TestLoadSo --leak-check=full -v
==26828== Memcheck, a memory error detector
==26828== Copyright (C) 2002-2015, and GNU GPL'd, by Julian Seward et al.
==26828== Using Valgrind-3.12.0 and LibVEX; rerun with -h for copyright info
==26828== Command: ./TestLoadSo --leak-check=full -v
==26828== 
!!!Hello World!!!
==26828== Jump to the invalid address stated on the next line
==26828==    at 0x0: ???
==26828==    by 0x53E63F0: (below main) (libc-start.c:291)
==26828==  Address 0x0 is not stack'd, malloc'd or (recently) free'd
==26828== 
==26828== 
==26828== Process terminating with default action of signal 11 (SIGSEGV)
==26828==  Bad permissions for mapped region at address 0x0
==26828==    at 0x0: ???
==26828==    by 0x53E63F0: (below main) (libc-start.c:291)
==26828== 
==26828== HEAP SUMMARY:
==26828==     in use at exit: 3,126 bytes in 9 blocks
==26828==   total heap usage: 13 allocs, 4 frees, 76,998 bytes allocated
==26828== 
==26828== LEAK SUMMARY:
==26828==    definitely lost: 0 bytes in 0 blocks
==26828==    indirectly lost: 0 bytes in 0 blocks
==26828==      possibly lost: 0 bytes in 0 blocks
==26828==    still reachable: 3,126 bytes in 9 blocks
==26828==         suppressed: 0 bytes in 0 blocks
==26828== Rerun with --leak-check=full to see details of leaked memory
==26828== 
==26828== For counts of detected and suppressed errors, rerun with: -v
==26828== ERROR SUMMARY: 1 errors from 1 contexts (suppressed: 0 from 0)
[1]    26828 segmentation fault (core dumped)  valgrind ./TestLoadSo --leak-check=full -v

C ++主类

extern "C" typedef char* (*helloWorld_t)();

int main() {

    void* handle = dlopen("./libMyLib.dll.so", RTLD_LAZY);

    if (!handle) {
     cerr << "Cannot open library: " << dlerror() << '\n';
     return 1;
     }
    helloWorld_t hello = (helloWorld_t)dlsym( handle, "helloWorld" );
    const char * tmp = hello();
     printf("\n%s",tmp);

    return 0;
}

extern功能是：

extern "C++" char* helloWorld() {
    char str[25];
    strcpy(str, "HelloWorld");
}

如果我使用extern "C"，我会收到编译错误：

error: conflicting declaration of ‘char* helloWorld()’ with ‘C’ linkage
 extern "C" char* helloWorld() {

我真的不清楚我哪里出错了。

Answer 1

函数不能同时具有C和C ++链接，函数指针类型必须与其目标函数的链接匹配。

您的简单名称不能dlsym extern "C++"功能。您必须在两种情况下都使用extern "C"（推荐），或者始终使用extern "C++"并将dlsym(handle, "helloWorld")中的字符串替换为您的函数的错位名称（不推荐）。

请务必检查dlsym的结果，如果返回空指针则报告错误（使用dlerror()为dlopen做的{。}}。

不要使用字符数组或指针来表示字符串。字符串的类型称为std::string。

最后但并非最不重要的是，始终使用-Wall -Werror进行编译，以便捕获不会实际返回值的非void函数等内容。

Answer 2

这里有很多问题：

extern "C++" char* helloWorld() {
    char str[25];
    strcpy(str, "HelloWorld");
}

它应该使用"C"链接。它应该返回一些东西。并且它将字符串复制到局部变量，因此值在返回时会丢失。可能

extern "C" char* helloWorld() {
    static char str[25]; // will keep its value accross calls, not thread safe
    return strcpy(str, "HelloWorld"); // return pointer to start of str
}

请注意，多个调用都返回相同的静态缓冲区。如果需要副本，则需要让调用者提供缓冲区，或者使用malloc分配的返回缓冲区。

在c ++中加载共享库会导致分段错误

2 个答案: